A361892 -id:A361892 - cp's OEIS frontend

A003162 A binomial coefficient summation.

1, 1, 1, 3, 6, 19, 49, 163, 472, 1626, 5034, 17769, 57474, 206487, 688881, 2508195, 8563020, 31504240, 109492960, 406214878, 1432030036, 5349255726, 19077934506, 71672186953, 258095737156, 974311431094, 3537275250214, 13408623649893

Offset: 0

N. J. A. Sloane

From Peter Bala, Mar 26 2023: (Start)
For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). For other cases see A361888 ({S(5,n)/S(1,n)}) and A361891 ({S(7,n)/ S(1,n)}).
Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^k) == b(n*p^(k-1)) (mod p^(3*k)) holds for positive integers n and k and all primes p >= 5. See A183069. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..1000
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161, A183069, A361887, A361888, A361889, A361890, A361891, A361892.

H := hypergeom([1/2,1/2],[1],16*x^2);
ogf := (Int(6*H*(4*x^2+5)/(4-x^2)^(3/2),x)+H*(16*x^2-1)/(4-x^2)^(1/2))*((2-x)/(2+x))^(1/2)/(4*x)+1/(8*x);
series(ogf,x=0,20);  # Mark van Hoeij, May 06 2013

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^3/Binomial[n, Floor[n/2]],{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)

a(n)=if(n<0, 0, sum(k=0,n\2, (binomial(n,k)-binomial(n,k-1))^3)/binomial(n,n\2)) /* Michael Somos, Jun 02 2005 */

G.f.: hypergeometric expression with an antiderivative, see Maple program. - Mark van Hoeij, May 06 2013
Recurrence: 4*n*(n+1)^2*(196*n^3 - 819*n^2 + 530*n + 528)*a(n) = 2*n*(1372*n^4 - 3633*n^3 - 7455*n^2 + 21934*n - 8448)*a(n-1) + (12740*n^6 - 90867*n^5 + 195310*n^4 - 13277*n^3 - 452690*n^2 + 528384*n - 174960)*a(n-2) + 8*(n-2)*(686*n^4 - 3010*n^3 + 1176*n^2 + 6543*n - 4725)*a(n-3) - 16*(n-3)^2*(n-2)*(196*n^3 - 231*n^2 - 520*n + 435)*a(n-4). - Vaclav Kotesovec, Mar 06 2014
a(n) ~ 4^(n+2)/(9*Pi*n^2). - Vaclav Kotesovec, Mar 06 2014

A361887 a(n) = S(5,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 2, 33, 276, 4150, 65300, 1083425, 20965000, 399876876, 8461219032, 178642861782, 4010820554664, 90684123972156, 2130950905378152, 50560833176021025, 1231721051614138800, 30294218438009039800, 759645100717216142000, 19213764100954274616908, 493269287121905287769776

Offset: 0

Peter Bala, Mar 28 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(5,n)}. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

a(n) is the total number of 5-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Seiichi Manyama, Table of n, a(n) for n = 0..673
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).
Column k=5 of A357824.
Cf. A008315, A120730.

seq(add( ( binomial(n,k) - binomial(n,k-1) )^5, k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^5, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)

from math import comb
def A361887(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**5 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^5.
a(n) = Sum_{j=0..n} A120730(n,j)^5.
a(n) = A357824(n,5). (End)
a(n) ~ 2^(5*n + 19/2) / (125 * Pi^(5/2) * n^(9/2)). - Vaclav Kotesovec, Aug 27 2023

A361890 a(n) = S(7,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 2, 129, 2316, 94510, 4939220, 211106945, 14879165560, 828070125876, 61472962084968, 4223017425122958, 325536754765395096, 25399546083773839692, 2059386837863675003112, 173281152533121109073025, 14789443838781868027714800, 1307994690673355979749969800

Offset: 0

Peter Bala, Mar 30 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(7,n)}. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

a(n) is the total number of 7-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Seiichi Manyama, Table of n, a(n) for n = 0..483
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/ S(1,2*n-1) ).
Column k=7 of A357824.
Cf. A008315, A120730.

seq(add( ( binomial(n,k) - binomial(n,k-1) )^7, k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^7, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)

from math import comb
def A361890(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^7.
a(n) = Sum_{j=0..n} A120730(n,j)^7.
a(n) = A357824(n,7). (End)
a(n) ~ 3 * 2^(7*n + 27/2) / (2401 * Pi^(7/2) * n^(13/2)). - Vaclav Kotesovec, Aug 27 2023

A361888 a(n) = S(5,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 1, 11, 46, 415, 3265, 30955, 299500, 3173626, 33576266, 386672861, 4340714886, 52846226091, 620906440961, 7857161332715, 95704821415240, 1246162831674580, 15624127945644100, 207990691516965886, 2669841775757784796, 36176886727828945286, 473508685502539872586

Offset: 0

Peter Bala, Mar 29 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). The present sequence is {S(5,n)/S(1,n)}.

Seiichi Manyama, Table of n, a(n) for n = 0..840
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(n,k) - binomial(n,k-1) )^5/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^5/Binomial[n, Floor[n/2]], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
a(n) = s(5, n)/s(1, n); \\ Seiichi Manyama, Mar 24 2025

a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.

a(n) ~ 2^(4*n + 9) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025

A361889 a(n) = S(5,2n-1)/S(1,2n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 11, 415, 30955, 3173626, 386672861, 52846226091, 7857161332715, 1246162831674580, 207990691516965886, 36176886727828945286, 6510211391453319830461, 1205449991704260042021490, 228686327051301858363357905, 44299708036441260810228742915, 8738765548899621077157770551275

Offset: 1

Peter Bala, Mar 29 2023

Odd bisection of A361888.

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

			Examples of supercongruences:
a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).
a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)

Seiichi Manyama, Table of n, a(n) for n = 1..420
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^5/binomial(2*n-1,n-1), k = 0..n-1), n = 1..20);

Table[Sum[(Binomial[2*n-1, k]-Binomial[2*n-1, k-1])^5 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

from math import comb
def A361889(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**5 for j in range(n))//comb((n<<1)-1,n-1) # Chai Wah Wu, Mar 25 2025

a(n) = 1/binomial(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n - k) * binomial(2*n-1,k) )^5 for n >= 1.

a(n) ~ 2^(8*n + 1) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025

A361891 a(n) = S(7,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 1, 43, 386, 9451, 246961, 6031627, 212559508, 6571985126, 243940325734, 9140730357409, 352312505157354, 14801600281919487, 600054439936968241, 26927918031565051915, 1149140935414286560040, 53804800109969394477580, 2401141625752684697505820

Offset: 0

Peter Bala, Mar 30 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). The present sequence is {S(7,n)/S(1,n)}.

Seiichi Manyama, Table of n, a(n) for n = 0..563
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(n,k) - binomial(n,k-1) )^7/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^7/Binomial[n, Floor[n/2]], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
a(n) = s(7, n)/s(1, n); \\ Seiichi Manyama, Mar 24 2025

from math import comb
def A361891(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1))//comb(n,n>>1) # Chai Wah Wu, Mar 25 2025

a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.

a(n) ~ 3 * 2^(6*n+13) / (2401 * Pi^3 * n^6). - Vaclav Kotesovec, Mar 24 2025

cp's OEIS Frontend

A003162 A binomial coefficient summation.

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A361887 a(n) = S(5,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

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A361890 a(n) = S(7,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

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A361888 a(n) = S(5,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

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A361889 a(n) = S(5,2*n-1)/S(1,2*n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

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A361891 a(n) = S(7,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

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A361889 a(n) = S(5,2n-1)/S(1,2n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.