A202917 -id:A202917 - cp's OEIS frontend

A203269 Row sums of triangle A202917.

1, 2, 8, 4, 132, 24, 1556, 88, 5364, 168, 5612, 136, 1128168, 28656, 86804, 1672, 3434920, 48080, 173886368, 2462176, 4093653652, 62126504, 4292224444, 10991336, 182512332988, 4948498616

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Dec 31 2011

nonn

If p is prime, then a(p-1)==2 mod p. At least up to n=251, there is only one nonprime n for which a(n-1)==2 mod n. It is n=18 for which a(18-1)=48080==2(mod 18).

Cf. A202917, A202941.

A175669 Triangle of numerators of coefficients of the polynomial Q^(2)m(n) defined by the recursion Q^(2)_0(n)=1; for m>=1, Q^(2)_m(n) = Sum{i=1..n} i^2Q^(2)_(m-1)(i). For m>=0, the denominator for all 3m+1 terms of the m-th row is A202367(m+1).

Original entry on oeis.org

1, 2, 3, 1, 0, 20, 96, 155, 90, 5, -6, 0, 280, 2772, 10518, 18711, 14385, 1323, -2863, -126, 360, 0, 2800, 47040, 323336, 1157760, 2238855, 2050020, 207158, -810600, -58505, 322740, 7956, -45360, 0, 12320, 314160, 3409472, 20401128, 72418826, 150057435, 154651321, 12413874, -101524412, -6408765, 82588957, 3394248, -37374084, -546480, 5443200, 0

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Dec 20 2011

Consider sequence of sequences of polynomials {Q^(0)_m(x)}, {Q^(1)_m(x)},...,{Q^(r)_m(x)},..., such that in every sequence m=0,1,...

Sequence {Q^(r)m(x)} is defined by the recursion: Q^(r)_0(x)=1; for m>=1 and integer x=n, Q^(r)_m(n)=sum{i=1,...,n}i^rQ^(r)(m-1)(i). By the induction, we see that polynomial Q^(r)_m(x) has degree (r+1)*m. Note that Q^(0)_m(n) is C(n+m-1,m), Q^(1)_m(n)=S(n+m,n), where S(k,l) are Stirling numbers of the second kind. Thus Q^(r)_m(x) is an r-generalization of binomial coefficients and Stirling numbers of the second kind. Moreover, for every r, LCM of denominators of the coefficients of Q^(r)_m(x) generate sequences of factorial type which possess important arithmetic properties. For r=0, it is n!, for r=1, it is A053657, for r=2,3,4 we obtain A202367, A202368, A202369. Denote the general term of the sequence corresponding to a given r by n!^(r) and, for 0<=m<=n, denote C^(r)(n,m)=n!^(r)/(m!^(r)*(n-m)!^(r). Then, for the "r-Pascal triangle", we have the following conjectural regularity: if a prime p==1 mod r, then the ((p-1)/r)-th row contains two 1's and numbers multiple of p. Cf. triangles A202917, A202941.

			The sequence of polynomials begins:
Q^(2)_0=1,
Q^(2)_1=(2*x^3+3*x^2+x)/6,
Q^(2)_2=(20*x^6+96*x^5+155*x^4+90*x^3+5*x^2-6*x)/360,
Q^(2)_3=(280*x^9+2772*x^8+10518*x^7+18711*x^6+14385*x^5+1323*x^4-2863*x^3 -126*x^2+360*x)/45360.

Cf. A202339, A053657, A202367, A202368, A202369.

Q^(2)_n(1)=1.

A202941 For n>=0, let n!^(2)=A202367(n+1) and, for 0<=m<=n, C^(2)(n,m)=n!^(2)/(m!^(2)*(n-m)!^(2)). The sequence gives triangle of numbers C^(2)(n,m) with rows of length n+1.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 21, 21, 1, 1, 20, 42, 20, 1, 1, 11, 22, 22, 11, 1, 1, 2730, 3003, 2860, 3003, 2730, 1, 1, 1, 273, 143, 143, 273, 1, 1

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Dec 26 2011

Conjecture. If p is an odd prime, then the ((p-1)/2)-th row contains two 1's and (p-3)/2 numbers multiple of p.

See also comments in A175669 and A202917.

			Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1....10.....1
.3..|..1....21 ...21.....1
.4..|..1....20....42....20.....1
.5..|..1....11....22....22....11.....1
.6..|..1..2730..3003..2860..3003..2730.....1
.7..|..1.....1...273...143...143...273.....1.....1
.8..|

Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917

If conjectural formula in A202367 is true, then A007814(C^(2)(n,m)) =A007814(C(n,m)).

A203278 Row sums of triangle A202941.

Original entry on oeis.org

1, 2, 12, 44, 84, 68, 14328, 836, 24040, 1231088, 31063252, 5495668, 2474249308, 4505692, 130544952, 2741953536, 6519198872, 11977808, 2197436928400, 5058240848, 11783977054420, 363453226568756, 135517101268864, 6345558675936, 67883449330825012, 1419845303575428

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Dec 31 2011

nonn

Conjecture. If p is odd prime, then a((p-1)/2)==2 (mod p)

Cf. A202941, A202917, A203269

A203484 For n>=0, let n!^(3) = A202368(n+1) and, for 0<=m<=n, C^(3)(n,m) = n!^(3)/(m!^(3)*(n-m)!^(3)). The sequence gives triangle of numbers C^(3)(n,m) with rows of length n+1.

Original entry on oeis.org

1, 1, 1, 1, 42, 1, 1, 5, 5, 1, 1, 1092, 130, 1092, 1, 1, 1, 26, 26, 1, 1, 1, 11970, 285, 62244, 285, 11970, 1, 1, 11, 3135, 627, 627, 3135, 11, 1

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 02 2012

Conjecture. If p is prime of the form 3*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 3*k+2, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.

			Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1......1
.2..|..1.....42.....1
.3..|..1......5 ....5......1
.4..|..1...1092...130...1092.....1
.5..|..1......1....26.....26.....1......1
.6..|..1..11970...285..62244...285..11970....1
.7..|..1.....11..3135....627...627...3135...11.....1
.8..|

Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941.

Conjecture. A007814(C^(3)(n,m)) = A007814(C(n,m)).

A178473 For n>=0, let n!^(4) = A202369(n+1) and, for 0<=m<=n, C^(4)(n,m) = n!^(4)/(m!^(4)*(n-m)!^(4)). The sequence gives triangle of numbers C^(4)(n,m) with rows of length n+1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 273, 273, 1, 1, 68, 9282, 68, 1, 1, 55, 1870, 1870, 55, 1, 1, 546, 15015, 3740, 15015, 546, 1, 1, 29, 7917, 1595, 1595, 7917, 29, 1

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 02 2012

Conjecture. If p is prime of the form 4*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 4*k+3, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.

			Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1......1
.2..|..1......2......1
.3..|..1....273 ...273......1
.4..|..1.....68...9282.....68......1
.5..|..1.....55...1870...1870.....55......1
.6..|..1....546..15015...3740..15015....546....1
.7..|..1.....29...7917...1595...1595...7917...29.....1
.8..|

Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941, A203484.

Conjecture. A007814(C^(4)(n,m)) = A007814(C(n,m)).

A204087 Reduced Pascal triangle: C_R(n,m) = A003418(n) / max(A003418(m), A003418(n-m)), m=0,...,n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 2, 6, 2, 1, 1, 5, 10, 10, 5, 1, 1, 1, 5, 10, 5, 1, 1, 1, 7, 7, 35, 35, 7, 7, 1, 1, 2, 14, 14, 70, 14, 14, 2, 1, 1, 3, 6, 42, 42, 42, 42, 6, 3, 1, 1, 1, 3, 6, 42, 42, 42, 6, 3, 1, 1, 1, 11, 11, 33, 66, 462, 462, 66, 33, 11, 11, 1

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 10 2012

The sixth row is the first one which differs from triangles A080381, A080396.

			Triangle begins:
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....1
.3..|..1.....3.....3.....1
.4..|..1.....2.....6.....2.....1
.5..|..1.....5....10....10.....5.....1
.6..|..1.....1.....5....10.....5.....1.....1
.7..|..1.....7.....7....35....35.....7.....7.....1

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A007318, A080381, A080396, A186430, A202917, A202941.

g:= proc(n) option remember; `if`(n=0, 1, ilcm(g(n-1), n)) end:
CR:= proc(n, m) option remember; g(n)/max(g(m), g(n-m)) end:
seq (seq (CR(n,m), m=0..n), n=0..11); # Alois P. Heinz, Jan 11 2012

g[n_] := g[n] = If[n == 0, 1, LCM[g[n-1], n]]; CR[n_, m_] := CR[n, m] = g[n]/Max[ g[m], g[n-m]]; Table[Table[CR[n, m], {m, 0, n}], {n, 0, 11}] // Flatten (* Jean-François Alcover, Mar 12 2015, after Alois P. Heinz *)

A203498 Row sums of triangle A203484.

Original entry on oeis.org

1, 2, 44, 12, 2316, 56, 86756, 7548, 2745504, 14216, 8303228, 27440, 26755926552, 17777912, 337554780656, 166149231952, 41032825702988, 10165065032, 19445273523788, 559223251080, 52853093762480872, 2217608618621076, 3130076262074284420, 9840817961344

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 02 2012

nonn

Conjecture. If prime p==1 (mod 3), then a((p-1)/3)==2 (mod p); if prime p==2 mod 3, then a((2*p-1)/3)==2 (mod p).

Cf. A202941, A202917, A203269, A203278, A203484, A178473.

A203509 Row sums of triangle A178473.

Original entry on oeis.org

1, 2, 4, 548, 9420, 3852, 34864, 19084, 15296, 154560176, 7180221844, 887102780, 211332046788, 71893259484, 20454788424, 5986072942766808, 5988933869570752, 22285488224, 2756032824242080, 17677170921656, 679436626785756, 20936052371230100988

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 02 2012

nonn

Conjecture. If prime p==1 (mod 4), then a((p-1)/4)==2 (mod p); if prime p==3 (mod 4), then a((p-1)/2)==2 (mod p).

Cf. A202941, A202917, A203269, A203278, A203484, A178473, A203498.

A204088 Row sums of A204087.

Original entry on oeis.org

1, 2, 4, 8, 12, 32, 24, 100, 132, 188, 148, 1168, 708, 3200, 2344, 1488, 2120, 21456, 16596, 222948, 130572, 38196, 29800, 492248, 299096, 529712, 455424, 1143404, 920536, 20232316, 13769088, 226481600, 252603072, 52242944, 40457056, 28671168, 16885280

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jan 10 2012

nonn

If n is prime, then a(n)==2 mod n. Up to n=200, there are only two composite numbers with such a property: a(49)=623017500436==2 mod 49 and a(51)=391496243228==2 mod 51.

Cf. A204087, A186430, A202917, A202941, A203269, A203278.

cp's OEIS Frontend

A203269 Row sums of triangle A202917.

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A175669 Triangle of numerators of coefficients of the polynomial Q^(2)m(n) defined by the recursion Q^(2)_0(n)=1; for m>=1, Q^(2)_m(n) = Sum{i=1..n} i^2*Q^(2)_(m-1)(i). For m>=0, the denominator for all 3*m+1 terms of the m-th row is A202367(m+1).

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A202941 For n>=0, let n!^(2)=A202367(n+1) and, for 0<=m<=n, C^(2)(n,m)=n!^(2)/(m!^(2)*(n-m)!^(2)). The sequence gives triangle of numbers C^(2)(n,m) with rows of length n+1.

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A203278 Row sums of triangle A202941.

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A203484 For n>=0, let n!^(3) = A202368(n+1) and, for 0<=m<=n, C^(3)(n,m) = n!^(3)/(m!^(3)*(n-m)!^(3)). The sequence gives triangle of numbers C^(3)(n,m) with rows of length n+1.

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A178473 For n>=0, let n!^(4) = A202369(n+1) and, for 0<=m<=n, C^(4)(n,m) = n!^(4)/(m!^(4)*(n-m)!^(4)). The sequence gives triangle of numbers C^(4)(n,m) with rows of length n+1.

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A204087 Reduced Pascal triangle: C_R(n,m) = A003418(n) / max(A003418(m), A003418(n-m)), m=0,...,n.

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A203498 Row sums of triangle A203484.

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A203509 Row sums of triangle A178473.

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A204088 Row sums of A204087.

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A175669 Triangle of numerators of coefficients of the polynomial Q^(2)m(n) defined by the recursion Q^(2)_0(n)=1; for m>=1, Q^(2)_m(n) = Sum{i=1..n} i^2Q^(2)_(m-1)(i). For m>=0, the denominator for all 3m+1 terms of the m-th row is A202367(m+1).