A000749 -id:A000749 - cp's OEIS frontend

A323100 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = -1 in Clifford algebra Cl(p,q)(R).

0, 0, 1, 1, 1, 3, 4, 2, 4, 6, 10, 6, 6, 10, 10, 20, 16, 12, 16, 20, 16, 36, 36, 28, 28, 36, 36, 28, 64, 72, 64, 56, 64, 72, 64, 56, 120, 136, 136, 120, 120, 136, 136, 120, 120, 240, 256, 272, 256, 240, 256, 272, 256, 240, 256, 496, 496, 528, 528, 496, 496, 528, 528, 496, 496, 528

Offset: 0

Jianing Song, Jan 04 2019

Cl(p,q)(R) is a 2^(p+q)-dimensional algebraic structure generated by {e_1, e_2, ..., e_(p+q)}, where (e_1)^2 = (e_2)^2 = ... = (e_p)^2 = +1, (e_(p+1))^2 = (e_(p+2))^2 = ... = (e_q)^2 = -1, (e_i)*(e_j) = -(e_j)*(e_i) for any i != j (anti-commutativity), ((e_i)*(e_j))*(e_k) = (e_i)*((e_j)*(e_k)) for any i, j, k (associativity). So the 2^(p+q) basis are all elements of the form Product_{s=1..t} e_(i_s) where 1 <= i_1 < i_2 < ... < i_t <= p + q and {i_1, i_2, ..., i_t} runs through all 2^(p+q) subsets of {1, 2, ..., p + q} (due to the failure of commutativity, one should be careful when taking continued products). Examples include: the real numbers Cl(0,0)(R), the complex numbers Cl(0,1)(R), split-complex numbers Cl(1,0)(R), quaternions Cl(1,0)(R), etc. If p + q = p' + q', then Cl(p,q)(R) is equal to Cl(p',q')(R) if and only if T(p,q) = T(p',q').
It can be shown that (Product_{s=1..t} e_(i_s))^2 = (-1)^(t*(t-1)/2)*(Product_{s=1..t} (e_(i_s))^2). So (Product_{s=1..t} e_(i_s))^2 = -1 if and only if t == 0, 1 (mod 4) and #({i_1, i_2, ..., i_t} intersect {p + 1, p + 2, ..., p + q}) is odd, or t == 2, 3 (mod 4) and #({i_1, i_2, ..., i_t} intersect {p + 1, p + 2, ..., p + q}) is even.
In general, let A = (a_ij) be any n X n symmetric {-1,1}-matrix, we can define an algebraic structure generated by {e_1, e_2, ..., e_n} where (e_i)^2 = a_ii for i = 1..n, (e_i)*(e_j) = (a_ij)*(e_j)*(e_i) for any i != j, ((e_i)*(e_j))*(e_k) = (e_i)*((e_j)*(e_k)) for any i, j, k. Clifford algebras are the cases where a_ij = -1 for any i != j. It can be shown that (Product_{s=1..t} e_(i_s)) * (Product_{s'=1..u} e_(j_s')) = (Product_{s=1..t, s'=1..u, i_s>=j_s'} a_(i_s)(j_s')) * (Product_{s=1..v} e_(k_s)), where 1 <= k_1 < k_2 < ... < k_v <= n and {k_1, k_2, ..., k_v} is the symmetric difference between {i_1, i_2, ..., i_t} and {j_1, j_2, ..., j_u}. Specially, (Product_{s=1..t} e_(i_s)))^2 = Product_{1<=s'<=s<=n} a_ss'. The 2^n basis, together with their additive inverses, form a group of order 2^(n+1) under multiplication, which is abelian if and only if a_ij = 1 for any i != j (in this case, it is isomorphic to (C_2)^(n+1) if a_ii = 1 for i = 1..n, and (C_2)^(n-1) X C_4 otherwise). The structure of this group can be complicated. For example, when n = 2, it can be isomorphic to either (C_2)^3, C_2 X C_4, C_2 X D_4 or Q_8.

			Table begins
p\q|  0   1   2    3    4    5  ...
---+-------------------------------
0  |  0,  1,  3,   6,  10,  16, ...
1  |  0,  1,  4,  10,  20,  36, ...
2  |  1,  2,  6,  16,  36,  72, ...
3  |  4,  6, 12,  28,  64, 136, ...
4  | 10, 16, 28,  56, 120, 256, ...
5  | 20, 36, 64, 120, 240, 496, ...
...
Example for T(1,3) = 10: (Start)
1^2 = 1;
(e_1)^2 = 1;
(e_2)^2 = -1;
(e_3)^2 = -1;
(e_4)^2 = -1;
((e_1)*(e_2))^2 = -(e_1)^2*(e_2)^2 = 1;
((e_1)*(e_3))^2 = -(e_1)^2*(e_3)^2 = 1;
((e_1)*(e_4))^2 = -(e_1)^2*(e_4)^2 = 1;
((e_2)*(e_3))^2 = -(e_2)^2*(e_3)^2 = -1;
((e_2)*(e_4))^2 = -(e_2)^2*(e_4)^2 = -1;
((e_3)*(e_4))^2 = -(e_3)^2*(e_4)^2 = -1;
((e_1)*(e_2)*(e_3))^2 = -(e_1)^2*(e_2)^2*(e_3)^2 = -1;
((e_1)*(e_2)*(e_4))^2 = -(e_1)^2*(e_2)^2*(e_4)^2 = -1;
((e_1)*(e_3)*(e_4))^2 = -(e_1)^2*(e_3)^2*(e_4)^2 = -1;
((e_2)*(e_3)*(e_4))^2 = -(e_2)^2*(e_3)^2*(e_4)^2 = 1;
((e_1)*(e_2)*(e_3)*(e_4))^2 = (e_1)^2*(e_2)^2*(e_3)^2*(e_4)^2 = -1. (End)
From _Peter Luschny_, Jan 13 2019: (Start)
The first few lines of the triangle T(i-j,j) are:
[0]   0;
[1]   0,   1;
[2]   1,   1,   3;
[3]   4,   2,   4,   6;
[4]  10,   6,   6,  10,  10;
[5]  20,  16,  12,  16,  20,  16;
[6]  36,  36,  28,  28,  36,  36,  28;
[7]  64,  72,  64,  56,  64,  72,  64,  56;
[8] 120, 136, 136, 120, 120, 136, 136, 120, 120;
[9] 240, 256, 272, 256, 240, 256, 272, 256, 240, 256; (End)

Jianing Song, Antidiagonals n = 0..99, flattened
Wikipedia, Clifford algebras

Cf. A038505(n+1) (first row), A000749(n+1) (first column), A006516 (main diagonal),
A321959 (antidiagonal sums).
A323346 is the complement sequence.

s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2,  s][1 + modp(n+1, 8)]:
T := proc(n, k) option remember;
if n = 0 then return 2^(k - 1) + 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^(n - 1) + 2^((n - 3)/2)*h(n) fi;
T(n, k-1 ) + T(n-1, k) end:
for n from 0 to 9 do seq(T(n, k), k=0..9) od; # Peter Luschny, Jan 12 2019

T[n_, k_] := Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];
Table[T[n-k, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 19 2019 *)

T(p,q) = sum(i=0, p, sum(j=0, q, binomial(p, i)*binomial(q, j)*(binomial(i-j, 2)%2)))

T(p,q) = Sum_{i=0..p} Sum_{j=0..q} binomial(p, i)*binomial(q, j)*(binomial(i - j, 2) mod 2).

T(p,q) = 2^(p+q) - A323346(p,q).

A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

Original entry on oeis.org

1, 4, 9, 16, 26, 44, 84, 176, 376, 784, 1584, 3136, 6176, 12224, 24384, 48896, 98176, 196864, 393984, 787456, 1573376, 3144704, 6288384, 12578816, 25163776, 50335744, 100675584, 201342976, 402661376, 805289984, 1610563584, 3221159936

Offset: 0

Creighton Dement, Nov 11 2004

A100215(n) (ves) = ((-1)^n)*A009116(n+3) (jes) + a(n) (les) + A038503(n+1) (tes) (Sn, below, corresponds to the generating function from above). Coefficients of Sn(z)*(1-z)/(1+z) gives match to A038504 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1"). Coefficients of Sn(z)/(1+z) gives match to A038505 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2"). Coefficients of Sn(z)/(1-z^2) gives match to A000749 (Number of strings over Z_2 of length n with trace 1 and subtrace 1). The elements 'i, 'j, 'k, i', j', k', 'ii', 'jj', 'kk', 'ij', 'ik', 'ji', 'jk', 'ki', 'kj', e ("floretions") are members of the quaternion product factor space Q x Q /{(1,1), (-1,-1)}. "les" sums over coefficients belonging to basis vectors which squared give the unit e (excluding e itself).

This sequence is identical to its 4th differences. - Jean-François Alcover, Nov 07 2013

			a(2) = 9 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 =
1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e
and the sum of the coefficients belonging to basis vectors which squared give the unit e (excluding e itself) is 3+2+2+1+1 = 9 (see comment).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000749, A009116, A009545, A038503, A038504, A038505, A100213, A100214, A100215.

[n le 3 select n^2 else 4*Self(n-1) -6*Self(n-2) +4*Self(n-3): n in [1..40]]; // G. C. Greubel, Mar 28 2024

a:= n-> (<<0|1|0>, <0|0|1>, <4|-6|4>>^n. <<1, 4, 9>>)[1, 1]:
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 07 2013

d = 4; nmax = 31; a[n_ /; n < d] := (n + 1)^2; seq = Table[a[n], {n, 0, nmax}]; seq /. Solve[ Thread[ Take[seq, nmax - d + 1] == Differences[seq, d]]] // First (* Jean-François Alcover, Nov 07 2013 *)
LinearRecurrence[{4,-6,4}, {1,4,9}, 41] (* G. C. Greubel, Mar 28 2024 *)

@CachedFunction
def a(n): # a = A100216
    if n<3: return (n+1)^2
    else: return 4*a(n-1) -6*a(n-2) +4*a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Mar 28 2024

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), with a(0) = 1, a(1) = 4, a(2) = 9.
G.f.: (1-x^2)/((1-2*x)*(1-2*x+2*x^2)).
(a(n)) = lesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).
2*a(n) = 3*2^n - A009545(n+1) + 4*A009545(n). - R. J. Mathar, May 21 2019
E.g.f.: (1/2)*exp(x)*(3*sin(x) - cos(x) + 3*exp(x)). - G. C. Greubel, Mar 28 2024

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

A130668 Diagonal of A129819.

Original entry on oeis.org

0, 0, 1, -2, 5, -11, 23, -48, 102, -220, 476, -1024, 2184, -4624, 9744, -20480, 42976, -90048, 188352, -393216, 819328, -1704192, 3539200, -7340032, 15203840, -31456256, 65010688, -134217728, 276826112, -570429440, 1174409216

Offset: 0

Paul Curtz, Jun 27 2007

sign

This sequence is connected to A124072. To see this, change the sign of every negative term and consider the differences of every line. Hence for the second line, and following lines, the four terms form periodic sequences:
 1   0   1   0
 0   0   1   1
 0   1   2   1
 1   3   3   1
 4   6   4   2
10  10   6   6
20  16  12  16
36  28  28  36
64  56  64  72
120 120 136 136
240 256 272 256.
The lines are connected as seen by the examples: (3rd line connected to 2nd, from right to left) 1+1=2, 1+0=1, 0+0=0, 0+1=1; (11th line connected to 10th) 136+136=272, 136+120=256, 120+120=240, 120+136=256.
The 4 columns are almost known (must the first line be suppressed?): A038503 (without the first 1), A000749 (without the first 0), A038505, A038504. Like the present sequence, every sequence of A124072 beginning with a negative number (-2, -11, ...) is a "twisted" sequence (see A129339 comments, A129961 and the present 4 columns). Periodic with period 2^n.
Inverse binomial transform of A129819. - R. J. Mathar, Feb 25 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-6,-14,-16,-8).

a:=[-2,5,-11,23];; for n in [5..30] do a[n]:=-6*a[n-1]+-14*a[n-2] -16*a[n-3]-8*a[n-4]; od; Concatenation([0,0,1], a); # G. C. Greubel, Mar 24 2019

I:=[-2,5,-11,23]; [0,0,1] cat [n le 4 select I[n] else -6*Self(n-1) - 14*Self(n-2)-16*Self(n-3)-8*Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 24 2019

gf = x^2*(1+x)*(1+3*x+4*x^2+3*x^3)/((1+2*x+2*x^2)*(1+2*x)^2); CoefficientList[Series[gf, {x, 0, 30}], x] (* Jean-François Alcover, Dec 16 2014, after R. J. Mathar *)
Join[{0, 0, 1}, LinearRecurrence[{-6,-14,-16,-8}, {-2,5,-11,23}, 30]] (* Jean-François Alcover, Feb 15 2016 *)

my(x='x+O('x^30)); concat([0,0], Vec(x^2*(1+x)*(1+3*x+4*x^2+3*x^3 )/((1+2*x +2*x^2)*(1+2*x)^2))) \\ G. C. Greubel, Mar 24 2019

(x^2*(1+x)*(1+3*x+4*x^2+3*x^3)/((1+2*x+2*x^2)*(1+2*x)^2 )).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 24 2019

From R. J. Mathar, Feb 25 2009: (Start)
G.f.: x^2*(1+x)*(1 + 3*x + 4*x^2 + 3*x^3)/((1 + 2*x + 2*x^2)*(1+2*x)^2).
a(n) = ((-1)^n*A001787(n+1) - 4*A108520(n) + 4*A122803(n))/32, n > 2. (End)
a(n) = -6*a(n-1) - 14*a(n-2) - 16*a(n-3) - 8*a(n-4) for n >= 7. - G. C. Greubel, Mar 24 2019

Extended by R. J. Mathar, Feb 25 2009

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

A085476 Periodic Pascal array, read by upward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 4, 3, 1, 1, 1, 1, 5, 6, 1, 2, 1, 1, 1, 6, 10, 4, 1, 1, 1, 1, 1, 7, 15, 10, 1, 3, 1, 1, 1, 1, 8, 21, 20, 5, 1, 3, 2, 1, 1, 1, 9, 28, 35, 15, 1, 4, 1, 1, 1, 1, 1, 10, 36, 56, 35, 6, 1, 6, 1, 1, 1, 1

Offset: 0

Paul Barry, Jul 02 2003

G.f. of binomial transform of n-th row is given by 1/((1-x)^(n+1)-x^(n+1)).

			Rows begin:
n\k | 0 1 2 3 4 5
----+------------
  0 | 1 1 1 1 1 1 ...
  1 | 1 1 1 1 1 1 ...
  2 | 1 2 1 1 2 1 ...
  3 | 1 3 3 1 1 3 ...
  4 | 1 4 6 4 1 1 ...

John Tyler Rascoe, Antidiagonals n = 0..139, flattened

Cf. A007318, A000749, A049016, A006090, A049017.

from math import comb
def A085476(n,k): return(comb(n,k%(n+1))) # John Tyler Rascoe, Dec 01 2023

Square array T(n, k) = C(n, k mod (n+1)).

A135360 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3) for n > 4, with first terms 1, 2, 4, 7.

Original entry on oeis.org

1, 2, 4, 7, 12, 22, 44, 92, 192, 392, 784, 1552, 3072, 6112, 12224, 24512, 49152, 98432, 196864, 393472, 786432, 1572352, 3144704, 6290432, 12582912, 25167872, 50335744, 100667392, 201326592, 402644992, 805289984, 1610596352, 3221225472, 6442483712, 12884967424

Offset: 0

Paul Curtz, Dec 08 2007

Sequence identical to its fourth differences.

Without a(3)=7, sequence A000079 would have been obtained. - Michel Marcus, May 06 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000079 (2^n), A000749.

Join[{1}, LinearRecurrence[{4, -6, 4}, {2, 4, 7}, 25]] (* G. C. Greubel, Oct 11 2016 *)

lista(nn) = {v = vector(nn); v[1] = 1; v[2] = 2; v[3] = 4; v[4] = 7; for (k=5, nn, v[k] = 4*v[k-1]-6*v[k-2]+4*v[k-3];); v;} \\ Michel Marcus, May 06 2015

a(n) = 2^n + A000749(n). - Michel Marcus, May 06 2015

G.f.: (1 - x)*(1 - x + x^2)/((1 - 2*x)*(1 - 2*x + 2*x^2)). [Bruno Berselli, May 06 2015]

More terms from Michel Marcus, May 06 2015

A132152 a(4n+k) = 4a(4n+k-1)-6a(4n+k-2)+4a(4n+k-3), for k = 0,1,2; 2*a(4n+3) = 7a(4n+2)-8(4n+1)+2a(4n), with a(0) = a(1) = a(2) = 0, a(3) = 1.

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 20, 34, 56, 100, 200, 356, 624, 1160, 2320, 4104, 7136, 13200, 26400, 46736, 81344, 150560, 301120, 533024, 927616, 1716800, 3433600, 6078016, 10577664, 19576960, 39153920, 69308544, 120618496, 223238400

Offset: 0

Paul Curtz, Nov 01 2007

a(n+1)-2a(n)= 0, 0, 1, 2, 2, 0, -6, -12, -12, 0, -44, -88, -88, 0

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 10, 0, 0, 0, 16).

Cf. A000749 (0, 0, 0, 1, 4, 10, 20, 36) for which a(n)=4a(n-1)-6a(n-2)+4a(n-3).

Join[{0},LinearRecurrence[{0,0,0,10,0,0,0,16},{0,0,1,4,10,20,34,56},40]] (* Harvey P. Dale, Nov 03 2013 *)

Sequence is identical to its fourth differences in absolute value.
a(n)=10*a(n-4)+16*a(n-8), n>8. - R. J. Mathar, Feb 07 2009
G.f.: -x^3*(2*x+1)*(4*x^2+1)*(2*x^2+2*x+1)/(-1+10*x^4+16*x^8) . - R. J. Mathar, Apr 19 2023

Definition corrected and the sequence extended by R. J. Mathar, Feb 07 2009

cp's OEIS Frontend

A323100 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = -1 in Clifford algebra Cl(p,q)(R).

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A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

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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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A130668 Diagonal of A129819.

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A135360 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3) for n > 4, with first terms 1, 2, 4, 7.