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when writing a program, that calculates pascals triangle, I noticed that the printout looks (to me) like an aproximately exponential curve

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1
1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1
1 20 190 1140 4845 15504 38760 77520 125970 167960 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1
1 21 210 1330 5985 20349 54264 116280 203490 293930 352716 352716 293930 203490 116280 54264 20349 5985 1330 210 21 1
1 22 231 1540 7315 26334 74613 170544 319770 497420 646646 705432 646646 497420 319770 170544 74613 26334 7315 1540 231 22 1
1 23 253 1771 8855 33649 100947 245157 490314 817190 1144066 1352078 1352078 1144066 817190 490314 245157 100947 33649 8855 1771 253 23 1
1 24 276 2024 10626 42504 134596 346104 735471 1307504 1961256 2496144 2704156 2496144 1961256 1307504 735471 346104 134596 42504 10626 2024 276 24 1
1 25 300 2300 12650 53130 177100 480700 1081575 2042975 3268760 4457400 5200300 5200300 4457400 3268760 2042975 1081575 480700 177100 53130 12650 2300 300 25 1
1 26 325 2600 14950 65780 230230 657800 1562275 3124550 5311735 7726160 9657700 10400600 9657700 7726160 5311735 3124550 1562275 657800 230230 65780 14950 2600 325 26 1
1 27 351 2925 17550 80730 296010 888030 2220075 4686825 8436285 13037895 17383860 20058300 20058300 17383860 13037895 8436285 4686825 2220075 888030 296010 80730 17550 2925 351 27 1
1 28 378 3276 20475 98280 376740 1184040 3108105 6906900 13123110 21474180 30421755 37442160 40116600 37442160 30421755 21474180 13123110 6906900 3108105 1184040 376740 98280 20475 3276 378 28 1
1 29 406 3654 23751 118755 475020 1560780 4292145 10015005 20030010 34597290 51895935 67863915 77558760 77558760 67863915 51895935 34597290 20030010 10015005 4292145 1560780 475020 118755 23751 3654 406 29 1
1 30 435 4060 27405 142506 593775 2035800 5852925 14307150 30045015 54627300 86493225 119759850 145422675 155117520 145422675 119759850 86493225 54627300 30045015 14307150 5852925 2035800 593775 142506 27405 4060 435 30 1
1 31 465 4495 31465 169911 736281 2629575 7888725 20160075 44352165 84672315 141120525 206253075 265182525 300540195 300540195 265182525 206253075 141120525 84672315 44352165 20160075 7888725 2629575 736281 169911 31465 4495 465 31 1
1 32 496 4960 35960 201376 906192 3365856 10518300 28048800 64512240 129024480 225792840 347373600 471435600 565722720 601080390 565722720 471435600 347373600 225792840 129024480 64512240 28048800 10518300 3365856 906192 201376 35960 4960 496 32 1
1 33 528 5456 40920 237336 1107568 4272048 13884156 38567100 92561040 193536720 354817320 573166440 818809200 1037158320 1166803110 1166803110 1037158320 818809200 573166440 354817320 193536720 92561040 38567100 13884156 4272048 1107568 237336 40920 5456 528 33 1

the number of numbers in each "level" of pascal's triangle grows linearly. The length of the biggest numbers grows proportional to the logarithm of a binomial coefficient which grows like the factorial function, soooo - I have no idea why, or if the length grows exponentially.

EDIT: dividing the lengths of each line by the length of the line before (without counting the space - the grow linearly) you get

2.0
1.5
1.3333333333333333
2.0
1.25
1.2
1.3333333333333333
1.3125
1.2380952380952381
1.1153846153846154
1.2413793103448276
1.1666666666666667
1.1428571428571428
1.1458333333333333
1.1272727272727272
1.0806451612903225
1.1044776119402986
1.162162162162162
1.0930232558139534
1.0638297872340425
1.12
1.0982142857142858
1.0731707317073171
1.0606060606060606
1.0857142857142856
1.0789473684210527
1.0731707317073171
1.0738636363636365
1.0582010582010581
1.065
1.0704225352112675

(this counts the number 1 as having the length 0)

calculating the $log_10$s without rounding you get similar results, although they are more constant. Anyway the qoutients seem to decrease, so it probably (no proof) doesn't grow exponentially, although I'd still like to know how this grows

I'm sorry if my English isn't very good.

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3
  • $\begingroup$ You could modify your program slightly to count the number of digits on each line. $\lceil \log_{10}n \rceil$ for each entry. $\endgroup$
    – Joffan
    Commented Jan 29, 2015 at 18:11
  • $\begingroup$ @Joffan hmm.. good idea $\endgroup$ Commented Jan 29, 2015 at 18:22
  • $\begingroup$ If you are calculating these as you go, you could work in logarithms in the first place to allow you to explore much larger numbers too. Binomial coefficients can be calculated using multiplication/division only so this is straightforward, eg ${33 \choose 3}=\frac{33!}{(33-3)!3!} = \frac{33\times 32\times 31}{3\times 2\times 1} = 5456$ $\endgroup$
    – Joffan
    Commented Jan 29, 2015 at 18:25

1 Answer 1

1
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The number of entries on the nth line is equal to n+1 (or n, depending on where you start counting).

The average number of digits on each line is approximately proportional n (actually n ln n;l see Stirling's approximation).

So the length of each line is approximately proportional to $n^2 ln(n)$.

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