A014068 a(n) = binomial(n*(n+1)/2, n).
1, 1, 3, 20, 210, 3003, 54264, 1184040, 30260340, 886163135, 29248649430, 1074082795968, 43430966148115, 1917283000904460, 91748617512913200, 4730523156632595024, 261429178502421685800, 15415916972482007401455, 966121413245991846673830, 64123483527473864490450300
Offset: 0
Keywords
Examples
From _Gus Wiseman_, Jan 25 2024: (Start) The a(0) = 1 through a(3) = 20 loop-graph edge-sets (loops shown as singletons): {} {{1}} {{1},{2}} {{1},{2},{3}} {{1},{1,2}} {{1},{2},{1,2}} {{2},{1,2}} {{1},{2},{1,3}} {{1},{2},{2,3}} {{1},{3},{1,2}} {{1},{3},{1,3}} {{1},{3},{2,3}} {{2},{3},{1,2}} {{2},{3},{1,3}} {{2},{3},{2,3}} {{1},{1,2},{1,3}} {{1},{1,2},{2,3}} {{1},{1,3},{2,3}} {{2},{1,2},{1,3}} {{2},{1,2},{2,3}} {{2},{1,3},{2,3}} {{3},{1,2},{1,3}} {{3},{1,2},{2,3}} {{3},{1,3},{2,3}} {{1,2},{1,3},{2,3}} (End)
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..370
- R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
- Eric Weisstein's World of Mathematics, Graph Loop.
Programs
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Magma
[Binomial(Binomial(n+1,2), n): n in [0..40]]; // G. C. Greubel, Feb 19 2022
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Mathematica
Binomial[First[#],Last[#]]&/@With[{nn=20},Thread[{Accumulate[ Range[ 0,nn]], Range[ 0,nn]}]] (* Harvey P. Dale, May 27 2014 *)
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Python
from math import comb def A014068(n): return comb(comb(n+1,2),n) # Chai Wah Wu, Jul 14 2024
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Sage
[(binomial(binomial(n+1, n-1), n)) for n in range(20)] # Zerinvary Lajos, Nov 30 2009
Formula
For n >= 1, Product_{k=1..n} a(k) = A022915(n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 08 2001
a(n) = binomial(T(n+1), T(n)) where T(n) = the n-th triangular number. - Amarnath Murthy, Jul 14 2005
a(n) = binomial(binomial(n+2, n), n+1) for n >= -1. - Zerinvary Lajos, Nov 30 2009
From Peter Bala, Feb 27 2020: (Start)
a(p) == (p + 1)/2 ( mod p^3 ) for prime p >= 5 (apply Mestrovic, equation 37).
Conjectural: a(2*p) == p*(2*p + 1) ( mod p^4 ) for prime p >= 5. (End)
a(n) = A084546(n,n). - Gus Wiseman, Jan 25 2024
a(n) = [x^n] (1+x)^(n*(n+1)/2). - Vaclav Kotesovec, Aug 06 2025
Comments