A139761 -id:A139761 - cp's OEIS frontend

A289389 a(n) = Sum_{k>=0} (-1)^kbinomial(n,5k+4).

0, 0, 0, 0, 1, 5, 15, 35, 70, 125, 200, 275, 275, 0, -1000, -3625, -9500, -21250, -42500, -76875, -124375, -171875, -171875, 0, 621875, 2250000, 5890625, 13171875, 26343750, 47656250, 77109375, 106562500, 106562500, 0, -385546875, -1394921875, -3651953125

Offset: 0

Vladimir Shevelev, Jul 05 2017

sign

{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively.

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5).

Cf. A139398, A133476, A139714, A139748, A139761.

Cf. A289306, A289321, A289387, A289388.

Table[Sum[(-1)^k*Binomial[n, 5 k + 4], {k, 0, n}], {n, 0, 36}] (* or *)
CoefficientList[Series[(-x^4)/((-1 + x)^5 - x^5), {x, 0, 36}], x] (* Michael De Vlieger, Jul 10 2017 *)

a(n) = sum(k=0, (n-4)\5, (-1)^k*binomial(n, 5*k+4)); \\ Michel Marcus, Jul 05 2017

G.f.: (-x^4)/((-1+x)^5 - x^5). - Peter J. C. Moses, Jul 05 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-8)/10) + (phi-1)^n*cos (3* Pi*(n-8)/10)), where phi is the golden ratio;
a(n+m) = a(n)*K_1(m) + K_4(n)*K_2(m) + K_3(n)*K_3(m) + K_2(n)*K_4(m) + K_1(n)*a(m), where K_1 is A289306, K_2 is A289321, K_3 is A289387, K_4 is A289388.
a(n) = 0 if and only if n=0,1,2 or n==3 (mod 10). - Vladimir Shevelev, Jul 15 2017

More terms from Peter J. C. Moses, Jul 05 2017

A101508 Product of binomial matrix and the Mobius matrix A051731.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 8, 4, 3, 1, 16, 8, 6, 4, 1, 32, 16, 11, 10, 5, 1, 64, 32, 21, 20, 15, 6, 1, 128, 64, 42, 36, 35, 21, 7, 1, 256, 128, 85, 64, 70, 56, 28, 8, 1, 512, 256, 171, 120, 127, 126, 84, 36, 9, 1, 1024, 512, 342, 240, 220, 252, 210, 120, 45, 10, 1, 2048, 1024, 683, 496, 385, 463, 462, 330, 165, 55, 11, 1

Offset: 0

Paul Barry, Dec 05 2004

Row sums are A101509. Diagonal sums are A101510.
The matrix inverse appears to be A128313. - R. J. Mathar, Mar 22 2013
Read as upper triangular matrix, this can be seen as "recurrences in A135356 applied to A023531" [Paul Curtz, Mar 03 2017]. - The columns are: A000079, A131577, A024495, A000749, A139761, ... Column n differs after the (n+1)-th nonzero term on from the binomial coefficients C(k,n). - M. F. Hasler, Mar 05 2017

			Rows begin
  1;
  2,1;
  4,2,1;
  8,4,3,1;
  16,8,6,4,1;
  ...

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

A101508 := proc(n,k)
    a := 0 ;
    for i from 0 to n do
        if modp(i+1,k+1) = 0 then
            a := a+binomial(n,i) ;
        end if;
    end do:
    return a;
end proc: # R. J. Mathar, Mar 22 2013

t[n_, k_] := Sum[If[Mod[i + 1, k + 1] == 0, Binomial[n, i], 0], {i, 0, n}]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 24 2014 *)

T(n,k)=sum(i=0,n, if((i+1)%(k+1)==0, binomial(n, i))) \\ M. F. Hasler, Mar 05 2017

T(n, k) = Sum_{i=0..n} if(mod(i+1, k+1)=0, binomial(n, i), 0).

Rows have g.f. x^k/((1-x)^(k+1)-x^(k+1)).

A289387 a(n) = Sum_{k>=0} (-1)^kbinomial(n, 5k+2).

Original entry on oeis.org

0, 0, 1, 3, 6, 10, 15, 20, 20, 0, -75, -275, -725, -1625, -3250, -5875, -9500, -13125, -13125, 0, 47500, 171875, 450000, 1006250, 2012500, 3640625, 5890625, 8140625, 8140625, 0, -29453125, -106562500, -278984375, -623828125, -1247656250, -2257031250

Offset: 0

Vladimir Shevelev, Jul 05 2017

{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the first Shevelev link respectively.
From Robert Israel, Jul 11 2017: (Start)
a(n)=0 for n == 9 (mod 10).
A112765(a(10*k)) = (5/2)*k - 3/4 - (-1)^k/4.
A112765(a(10*k+2)) = (5/2)*k - 1/4 + (-1)^k/4.
A112765(a(10*k+3)) = A112765(a(10*k+4)) = (5/2)*k + 1/4 - (-1)^k/4.
A112765(a(10*k+5)) = A112765(a(10*k+6)) = (5/2)*k + 3/4 + (-1)^k/4.
A112765(a(10*k+7)) = A112765(a(10*k+8)) = (5/2)*k + 5/4 - (-1)^k/4. (End)
Note that from author's formula (see below) we have that, except for zeros in the sequence mentioned by Robert Israel, there are only a(0) = a(1) = 0. Indeed, otherwise for some value of n we should have the equality (phi-1)^n = -cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10). However, the absolute value of the right hand side takes the six distinct values only: 1, phi, phi^2, phi^(-1), phi^(-2), 1/3 (the last value we have when n == 9 (mod 10), since lim_{x->Pi/2}cos(x)/cos(3*x)= -1/3). Thus for n>=3, we have (phi-1)^n = phi^(-n) < |cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10)|. - Vladimir Shevelev, Jul 15 2017

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Vladimir Shevelev, A matrix property of sums of binomial coefficients, Seqfan, Fri Jul 21 2017.
Wikipedia, Circulant matrix

Cf. A139398, A133476, A139714, A139748, A139761, A289306, A289321, A289388, A289389.

f:= gfun:-rectoproc({5*a(n)-10*a(n+1)+10*a(n+2)-5*a(n+3)+a(n+4), a(0)=0,
a(1)=0, a(2)=1, a(3) = 3,a(4)=6},a(n),remember):
map(f, [$0..40]); # Robert Israel, Jul 11 2017

Table[Sum[(-1)^k*Binomial[n, 5 k + 2], {k, 0, n}], {n, 0, 35}] (* or *)
CoefficientList[Series[-((-1 + x)^2 x^2)/((-1 + x)^5 - x^5), {x, 0, 35}], x] (* Michael De Vlieger, Jul 10 2017 *)

a(n) = sum(k=0, (n-2)\5, (-1)^k*binomial(n, 5*k+2)); \\ Michel Marcus, Jul 05 2017

G.f.: -((-1+x)^2*x^2)/((-1+x)^5 - x^5). - Peter J. C. Moses, Jul 05 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-4)/10) + (phi-1)^n*cos(3* Pi*(n-4)/10)), where phi is the golden ratio.
a(n+m) = a(n)*K_1(m) + K_2(n)*K_2(m) + K_1(n)*a(m) - K_5(n)*K_4(m) - K_4(n)*K_5(m), where K_1 is A289306, K_2 is A289321, K_4 is A289388, K_5 is A289389.
For every n>=1, the determinant of circulant matrix of order 5 (see [Wikipedia]) with the entries (-1)^(i-1)* K_i(n), i=1..5, is 0. Here K_1, K_2, K_4 and K_5 are the same as in the previous formula, while K_3(n) = a(n). For a proof and a generalization see the second Shevelev link that also contains two unsolved problems. - Vladimir Shevelev, Jul 26 2017

More terms from Peter J. C. Moses, Jul 05 2017

A157823 a(n) = A156591(n) + A156591(n+1).

Original entry on oeis.org

-5, -1, -2, -4, -8, -16, -32, -64, -128, -256, -512, -1024, -2048, -4096, -8192, -16384, -32768, -65536, -131072, -262144, -524288, -1048576, -2097152, -4194304, -8388608, -16777216, -33554432, -67108864, -134217728, -268435456, -536870912, -1073741824

Offset: 0

Paul Curtz, Mar 07 2009

A156591 = 2,-7,6,-8,4,-12,... a(n) is companion to A154589 = 4,-1,-2,-4,-8,.For this kind ,companion of sequence b(n) is first differences a(n), second differences being b(n). Well known case: A131577 and A011782. a(n)+b(n)=A000079 or -A000079. a(n)=A154570(n+2)-A154570(n) ,A154570 = 1,3,-4,2,-6,-2,-14,. See sequence(s) identical to its p-th differences (A130785,A130781,A024495,A000749,A138112(linked to Fibonacci),A139761).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Vec(-(9*x-5)/(2*x-1) + O(x^100)) \\ Colin Barker, Feb 03 2015

a(n) = 2*a(n-1) for n>1. G.f.: -(9*x-5) / (2*x-1). - Colin Barker, Feb 03 2015

Edited by Charles R Greathouse IV, Oct 11 2009

A290993 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^6.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 463, 804, 1365, 2366, 4368, 8736, 18565, 40410, 87381, 184604, 379050, 758100, 1486675, 2884776, 5592405, 10919090, 21572460, 43144920, 87087001, 176565486, 357913941, 723002336, 1453179126, 2906358252, 5791193143

Offset: 0

Clark Kimberling, Aug 21 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291000 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6).

Cf. A000012, A033453, A192080, A289780, A290991, A290992, A291000.

Sequences of the form x^(m-1)/((1-x)^m - x^m): A000079 (m=1), A131577 (m=2), A024495 (m=3), A000749 (m=4), A139761 (m=5), this sequence (m=6), A290994 (m=7), A290995 (m=8).

a:=[0,0,0,0,1];;  for n in [6..35] do a[n]:=6*a[n-1]-15*a[n-2]+20*a[n-3]-15*a[n-4]+6*a[n-5]; od; Concatenation([0],a); # Muniru A Asiru, Oct 23 2018

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0] cat Coefficients(R!( x^5/((1-x)^6 - x^6) )); // G. C. Greubel, Apr 11 2023

seq(coeff(series(x^5/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 23 2018

z = 60; s = x/(1 - x); p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290993 *)

concat(vector(5), Vec(x^5 / ((1 - 2*x)*(1 - x + x^2)*(1 - 3*x + 3*x^2)) + O(x^50))) \\ Colin Barker, Aug 24 2017

def A290993_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^5/((1-x)^6 - x^6) ).list()
A290993_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) for n>5. Corrected by Colin Barker, Aug 24 2017
G.f.: x^5 / ((1 - 2*x)*(1 - x + x^2)*(1 - 3*x + 3*x^2)). - Colin Barker, Aug 24 2017
a(n) = A192080(n-5) for n > 5. - Georg Fischer, Oct 23 2018
G.f.: x^5/((1-x)^6 - x^6). - G. C. Greubel, Apr 11 2023

A290994 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^7.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 7, 28, 84, 210, 462, 924, 1717, 3017, 5110, 8568, 14756, 27132, 54264, 116281, 257775, 572264, 1246784, 2641366, 5430530, 10861060, 21242341, 40927033, 78354346, 150402700, 291693136, 574274008, 1148548016, 2326683921, 4749439975

Offset: 0

Clark Kimberling, Aug 22 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291000 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,2).

Cf. A000012, A033453, A049017, A289780, A291000.

Sequences of the form x^(m-1)/((1-x)^m - x^m): A000079 (m=1), A131577 (m=2), A024495 (m=3), A000749 (m=4), A139761 (m=5), A290993 (m=6), this sequence (m=7), A290995 (m=8).

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0,0] cat Coefficients(R!( x^6/((1-x)^7 - x^7) )); // G. C. Greubel, Apr 11 2023

z = 60; s = x/(1 - x); p = 1 - s^7;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290994 *)

concat(vector(6), Vec(x^6 / ((1 - 2*x)*(1 - 5*x + 11*x^2 - 13*x^3 + 9*x^4 - 3*x^5 + x^6)) + O(x^50))) \\ Colin Barker, Aug 22 2017

def A290994_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^6/((1-x)^7 - x^7) ).list()
A290994_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + 2*a(n-7) for n >= 8.
G.f.: x^6 / ((1 - 2*x)*(1 - 5*x + 11*x^2 - 13*x^3 + 9*x^4 - 3*x^5 + x^6)). - Colin Barker, Aug 22 2017
a(n) = A049017(n-6) for n > 5. - Georg Fischer, Oct 23 2018
G.f.: x^6/((1-x)^7 - x^7). - G. C. Greubel, Apr 11 2023

A220755 Numbers n such that n^2 + n(n+1)/2 is an oblong number (A002378).

Original entry on oeis.org

0, 1, 28, 117, 2760, 11481, 270468, 1125037, 26503120, 110242161, 2597035308, 10802606757, 254482957080, 1058545220041, 24936732758548, 103726628957277, 2443545327380640, 10164151092593121, 239442505350544188, 995983080445168597, 23462921979025949800

Offset: 1

Alex Ratushnyak, Apr 13 2013

Numbers n such that 6*n^2 + 2*n + 1 is a square. - Joerg Arndt, Apr 14 2013
a(n+4) - a(n) is divisible by 40. (a(n+2) - a(n)) mod 10 = period 4: repeat 8, 6, 2, 4. See A000689. - Paul Curtz, Apr 15 2013
For this 5 consecutive terms recurrence,the main (or principal) sequence is: CRR(n)= 0, 0, 0, 0, 1, 1, 99, 99, 9702, 9702,... . - Paul Curtz, Apr 16 2013
Also numbers n such that the sum of the octagonal numbers N(n) and N(n+1) is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 09 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000217, A005449 (n^2 + n(n+1)/2).
Cf. A011916 (numbers n>=0 such that n^2 + n(n+1)/2 is a triangular number).
Cf. A220186 (numbers n>=0 such that n^2 + n(n+1)/2 is a square).
Cf. A220185 (numbers n>=0 such that n^2 + n(n+1) is an oblong number).
(Example of a family of main sequences: A131577, A024495, A000749, A139761. )
Cf. A251793.

#include 
typedef unsigned long long U64;
U64 rootPronic(U64 a) {
    U64 sr = 1L<<31, s, b;
    while (a < sr*(sr+1))  sr>>=1;
    for (b = sr>>1; b; b>>=1) {
            s = sr+b;
            if (a >= s*(s+1))  sr = s;
    }
    return sr;
}
int main() {
  U64 a, n, r, t;
  for (n=0; n < 3L<<30; n++) {
    a = n*(n+1)/2 + n*n;
    t = rootPronic(a);
    if (a == t*(t+1)) {
        printf("%llu\n", n);
    }
  }
}

LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 28, 117, 2760}, 30] (* Giovanni Resta, Apr 14 2013 *)
CoefficientList[Series[x (1 + 27 x - 9 x^2 - 3 x^3)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 13 2014 *)

makelist(expand(((-(-1)^n+sqrt(6))*(5+2*sqrt(6))^(n-1)-((-1)^n+sqrt(6))*(5-2*sqrt(6))^(n-1)-2)/12), n, 1, 25); /* Bruno Berselli, Apr 14 2013 */

concat([0], Vec( x * (1+27*x-9*x^2-3*x^3) / ( (1-x)*(1-10*x+x^2)*(1+10*x+x^2) ) + O(x^66) ) )  /* Joerg Arndt, Apr 14 2013 */

G.f.: x^2 * (1+27*x-9*x^2-3*x^3) / ( (1-x)*(1-10*x+x^2)*(1+10*x+x^2) ). - Giovanni Resta, Apr 14 2013, adapted by Vincenzo Librandi Aug 13 2014
a(n) = ((-(-1)^n+sqrt(6))*(5+2*sqrt(6))^(n-1)-((-1)^n+sqrt(6))*(5-2*sqrt(6))^(n-1)-2)/12. - Bruno Berselli, Apr 14 2013
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).

a(11)-a(21) from Giovanni Resta, Apr 14 2013

cp's OEIS Frontend

A289389 a(n) = Sum_{k>=0} (-1)^k*binomial(n,5*k+4).

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A101508 Product of binomial matrix and the Mobius matrix A051731.

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A289387 a(n) = Sum_{k>=0} (-1)^k*binomial(n, 5*k+2).

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A157823 a(n) = A156591(n) + A156591(n+1).

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A290993 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^6.

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A290994 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^7.

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A220755 Numbers n such that n^2 + n(n+1)/2 is an oblong number (A002378).

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A289389 a(n) = Sum_{k>=0} (-1)^kbinomial(n,5k+4).

A289387 a(n) = Sum_{k>=0} (-1)^kbinomial(n, 5k+2).