A006046 -id:A006046 - cp's OEIS frontend

A268526 a(n) = ra(ceiling(n/2))+sa(floor(n/2)) with a(1)=1 and (r,s)=(3,2).

1, 5, 17, 25, 61, 85, 109, 125, 233, 305, 377, 425, 497, 545, 593, 625, 949, 1165, 1381, 1525, 1741, 1885, 2029, 2125, 2341, 2485, 2629, 2725, 2869, 2965, 3061, 3125, 4097, 4745, 5393, 5825, 6473, 6905, 7337, 7625, 8273, 8705, 9137, 9425, 9857, 10145, 10433, 10625, 11273, 11705, 12137, 12425

Offset: 1

N. J. A. Sloane, Feb 16 2016

nonn

K.-N. Chang and S.-C. Tsai, Exact solution of a minimal recurrence, Inform. Process. Lett. 75 (2000), 61-64.

Sequences of form a(n) = r*a(ceiling(n/2))+s*a(floor(n/2)) with a(1)=1 and (r,s) = (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1): A000027, A006046, A064194, A130665, A073121, A268524, A116520, A268525, A268526, A268527.

[n le 1 select 1 else 3*Self(Ceiling(n/2))+2*Self(Floor(n/2)): n in [1..60]]; // Vincenzo Librandi, Aug 30 2016

a(n) = if (n==1, 1, 3*a(ceil(n/2))+2*a(floor(n/2))); \\ Michel Marcus, Aug 30 2016

A268527 a(n) = ra(ceiling(n/2))+sa(floor(n/2)) with a(1)=1 and (r,s)=(4,1).

Original entry on oeis.org

1, 5, 21, 25, 89, 105, 121, 125, 381, 445, 509, 525, 589, 605, 621, 625, 1649, 1905, 2161, 2225, 2481, 2545, 2609, 2625, 2881, 2945, 3009, 3025, 3089, 3105, 3121, 3125, 7221, 8245, 9269, 9525, 10549, 10805, 11061, 11125, 12149, 12405, 12661, 12725, 12981, 13045, 13109, 13125, 14149, 14405

Offset: 1

N. J. A. Sloane, Feb 16 2016

nonn

K.-N. Chang and S.-C. Tsai, Exact solution of a minimal recurrence, Inform. Process. Lett. 75 (2000), 61-64.

Sequences of form a(n) = r*a(ceiling(n/2))+s*a(floor(n/2)) with a(1)=1 and (r,s) = (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1): A000027, A006046, A064194, A130665, A073121, A268524, A116520, A268525, A268526, A268527.

a(n) = if (n==1, 1, 4*a(ceil(n/2))+a(floor(n/2))); \\ Michel Marcus, Aug 30 2016

A382731 Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 8.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 41, 51, 60, 72, 83, 97, 111, 127, 132, 142, 155, 175, 188, 206, 226, 250, 261, 283, 303, 331, 353, 381, 409, 441, 446, 456, 469, 489, 506, 532, 560, 600, 613, 639, 665, 701, 729, 769, 809, 857, 868, 890, 918, 962, 990, 1030, 1074, 1130, 1152, 1196, 1236, 1292, 1336, 1392, 1448, 1512, 1517, 1527

Offset: 0

N. J. A. Sloane, Apr 23 2025

nonn

James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 8), European Journal of Combinatorics 19:1 (1998), pp. 45-62.

Cf. A001316, A006046-A006048, A194458, A194459, A382720-A382730.

def A382731(n):
    c = 0
    for m in range(n+1):
        n1 = m>>1
        n2 = n1>>1
        np = ~m
        n100 = (n2&(~n1)&np).bit_count()
        n110 = (n2&n1&np).bit_count()
        n10 = (n1&np).bit_count()
        c += ((n100+1<<3)+(n110<<1)+n10*(n10+3))<>3
    return c # Chai Wah Wu, Aug 10 2025

A116522 a(0)=1, a(1)=1, a(n)=7a(n/2) for n=2,4,6,..., a(n)=6a((n-1)/2)+a((n+1)/2) for n=3,5,7,....

Original entry on oeis.org

0, 1, 7, 13, 49, 55, 91, 127, 343, 349, 385, 421, 637, 673, 889, 1105, 2401, 2407, 2443, 2479, 2695, 2731, 2947, 3163, 4459, 4495, 4711, 4927, 6223, 6439, 7735, 9031, 16807, 16813, 16849, 16885, 17101, 17137, 17353, 17569, 18865, 18901, 19117, 19333

Offset: 0

Roger L. Bagula, Mar 15 2006

nonn

A 7-divide version of A084230.
The Harborth: f(2^k) = 3^k suggests that a family of sequences of the form: f(2^k) = prime(n)^k.
From Gary W. Adamson, Aug 27 2016: (Start)
Let M = the production matrix below. Then lim_{k->infinity} M^k generates the sequence with offset 1 by extracting the left-shifted vector.
  1, 0, 0, 0, 0, ...
  7, 0, 0, 0, 0, ...
  6, 1, 0, 0, 0, ...
  0, 7, 0, 0, 0, ...
  0, 6, 1, 0, 0, ...
  0, 0, 7, 0, 0, ...
  0, 0, 6, 1, 0, ...
  ...
The sequence divided by its aerated variant is (1, 7, 6, 0, 0, 0, ...). (End)

Alois P. Heinz, Table of n, a(n) for n = 0..16383 (first 2501 terms from G. C. Greubel)
H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62, 19-22, 1977.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 27, 33.
Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant

Cf. A000120, A006046, A077465, A130665, A116520, A130667, A161342, A360189.

a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 7*a(n/2) else 6*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n),n=0..47);
# second Maple program:
b:= proc(n) option remember; `if`(n<0, 0,
      b(n-1)+x^add(i, i=Bits[Split](n)))
    end:
a:= n-> subs(x=6, b(n-1)):
seq(a(n), n=0..44);  # Alois P. Heinz, Mar 06 2023

b[0] := 0; b[1] := 1; b[n_?EvenQ] := b[n] = 7*b[n/2]; b[n_?OddQ] := b[n] = 6*b[(n - 1)/2] + b[(n + 1)/2]; a = Table[b[n], {n, 1, 25}]

G.f.: (r(x) * r(x^2) * r(x^4) * r(x^8) * ...), where r(x) = (1 + 7x + 6x^2).
a(n) = Sum_{k=0..n-1} 6^wt(k), where wt = A000120. - Mike Warburton, Mar 14 2019
a(n) = Sum_{k=0..floor(log_2(n))} 6^k*A360189(n-1,k). - Alois P. Heinz, Mar 06 2023

Edited by N. J. A. Sloane, Apr 16 2005

A134659 Total number of odd coefficients in (1+x+x^2)^k for k=0,...,n.

Original entry on oeis.org

1, 4, 7, 12, 15, 24, 29, 40, 43, 52, 61, 76, 81, 96, 107, 128, 131, 140, 149, 164, 173, 200, 215, 248, 253, 268, 283, 308, 319, 352, 373, 416, 419, 428, 437, 452, 461, 488, 503, 536, 545, 572, 599, 644, 659, 704, 737, 800, 805, 820, 835, 860, 875, 920, 945, 1000

Offset: 0

Steven Finch, Jan 25 2008

nonn

a(n) = Sum_{k <= n} A071053(k)

N. J. A. Sloane, Table of n, a(n) for n = 0..16383
S. R. Finch, P. Sebah and Z.-Q. Bai, Odd Entries in Pascal's Trinomial Triangle (arXiv:0802.2654)
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168, 2015

Cf. A001316, A006046, A027907, A071053.

Sum[PolynomialMod[(1+x+x^2)^k, 2] /. x->1, {k, 0, n-1}]

Offset changed to 0 by N. J. A. Sloane, Feb 06 2015

A231348 Number of triangles after n-th stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 23, 33, 41, 45, 53, 65, 81, 91, 111, 133, 149, 153, 161, 173, 189, 201, 225, 253, 285, 295, 315, 343, 383, 405, 449, 495, 527, 531, 539, 551, 567, 579, 603, 631, 663, 675, 699, 731, 779, 807, 863, 923, 987, 997, 1017, 1045, 1085, 1113, 1169, 1233, 1313, 1335, 1379, 1439, 1527, 1573, 1665, 1759, 1823

Offset: 0

Omar E. Pol, Dec 15 2013

nonn

On the semi-infinite square grid the structure of this C.A. contains "black" triangles and "gray" triangles (see the Links section). Both types of triangles have two sides of length 5^(1/2). Every black triangle has a base of length 2 hence its height is 2 and its area is 2. Every gray triangle has a base of length 2^(1/2) hence its height is 3/(2^(1/2)) and its area is 3/2. Both types of triangles are arranged in the same way as the triangles of Sierpinski gasket (see A047999 and A006046). The black triangles are arranged in vertical direction. On the other hand the gray triangles are arranged in diagonal direction in the holes of the structure formed by the black triangles. Note that the vertices of all triangles coincide with the grid points.
The sequence gives the total number of triangles (black and gray) in the structure after n-th stage. A231349 (the first differences) gives the number of triangles added at n-th stage.
For a more complex structure see A233780.

			We start at stage 0 with no triangles, so a(0) = 0.
At stage 1 we add a black triangle, so a(1) = 1.
At stage 2 we add two black triangles, so a(2) = 1+2 = 3.
At stage 3 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(3) = 3+2+2 = 7.
At stage 4 we add four black triangles, so a(4) = 7+4 = 11.
At stage 5 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(5) = 11+2+2 = 15.
At stage 6 we add four black triangles and four gray triangles, so a(6) = 15+4+4 = 23.
At stage 7 we add four black triangles and six gray triangles, so a(7) = 23+4+6 = 33.
At stage 8 we add eight black triangles, so a(8) = 33+8 = 41.
And so on.
Note that always we add both black triangles and gray triangles except if n is a power of 2. In this case at stage 2^k we add only 2^k black triangles, for k >= 0.

Omar E. Pol, Illustration of the structure after 16 stages
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS

Cf. A047999, A006046, A139250, A151566, A160406, A173530, A182634, A194444, A220524, A220478, A230981, A231349, A233780.

A014370 If n = binomial(b,2) + binomial(c,1), b > c >= 0 then a(n) = binomial(b+1,3) + binomial(c+1,2).

Original entry on oeis.org

1, 2, 4, 5, 7, 10, 11, 13, 16, 20, 21, 23, 26, 30, 35, 36, 38, 41, 45, 50, 56, 57, 59, 62, 66, 71, 77, 84, 85, 87, 90, 94, 99, 105, 112, 120, 121, 123, 126, 130, 135, 141, 148, 156, 165, 166, 168, 171, 175, 180, 186, 193, 201, 210, 220, 221, 223, 226, 230, 235, 241, 248, 256, 265, 275, 286

Offset: 1

N. J. A. Sloane

			The triangle starts:
  1
  2 4
  5 7 10
  11 13 16 20
  21 23 26 30 35

W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge, 1993, p. 159.

Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of triangle, flattened).

Cf. A002260, A000292 (main diagonal), A000217, A014368, A014369, A006046, A050407 (1st column), A005581 (subdiagonal), A071239 (row sums), A212013.

a := 0: for i from 1 to 15 do for j from 1 to i do a := a+j: printf(`%d,`,a); od:od:

A014370[n_, k_] := Binomial[n + 1, 3] + Binomial[k + 1, 2];
Table[A014370[n, k], {n, 12}, {k, n}] (* Paolo Xausa, Mar 11 2025 *)

a(n) = Sum_{m = 1..n} b(m), b(m) = 1,1,2,1,2,3,1,2,3,4,... = A002260. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
a(n*(n+1)/2+m) = n*(n+1)*(n+2)/6 + m*(m+1)/2 = A000292(n)+ A000217(m), m = 0...n+1, n = 1, 2, 3.. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
a(n) = a(n-1) + A002260(n). As a triangle with n >= k >= 1: a(n, k) = a(n-1, k) + (n-1)*n/2 = a(n, k-1) + k = (n^3-n+3k^2+3k)/6. - Henry Bottomley, Nov 14 2001
a(n) = b(n) * (b(n) + 1) * (b(n) + 2) / 6 + c(n) * (c(n) + 1) / 2, where b(n) = [sqrt(2 * n) - 1/2] and c(n) = n - b(n) * (b(n) + 1) / 2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 20 2002
As a triangle, T(n,k) = binomial(n+1, 3) + binomial(k+1,2). - Franklin T. Adams-Watters, Jan 27 2014

More terms from James Sellers, Feb 05 2000

A074330 a(n) = Sum_{k=1..n} 2^b(k) where b(k) denotes the number of 1's in the binary representation of k.

Original entry on oeis.org

2, 4, 8, 10, 14, 18, 26, 28, 32, 36, 44, 48, 56, 64, 80, 82, 86, 90, 98, 102, 110, 118, 134, 138, 146, 154, 170, 178, 194, 210, 242, 244, 248, 252, 260, 264, 272, 280, 296, 300, 308, 316, 332, 340, 356, 372, 404, 408, 416, 424, 440, 448, 464, 480, 512, 520, 536

Offset: 1

Benoit Cloitre, Oct 06 2002

Robert Israel, Table of n, a(n) for n = 1..10000
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 29.

a(n) = A006046(n+1)-1. Cf. A080263.

f:= proc(n) option remember; if n::even then 2*procname(n/2-1)+procname(n/2)+2
  else 3*procname((n-1)/2)+2
  fi
end proc:
f(0):= 0:
map(f, [$1..100]); # Robert Israel, Oct 08 2020

b[n_] := IntegerDigits[n, 2] // Total;
a[n_] := 2^(b /@ Range[n]) // Total;
Array[a, 100] (* Jean-François Alcover, Aug 16 2022 *)

a(n)=sum(i=1,n,2^sum(k=1,length(binary(i)), component(binary(i),k)))

a(n+1)-a(n) = A001316(n)
From Ralf Stephan, Oct 07 2003: (Start)
a(0)=0, a(2n) = 2a(n-1) + a(n) + 2, a(2n+1) = 3a(n) + 2.
G.f.: (1/(1-x)) * Product_{k>=0} (1 + 2x^2^k). (End)

A116525 a(0)=1, a(1)=1, a(n) = 11a(n/2) for even n, and a(n) = 10a((n-1)/2) + a((n+1)/2) for odd n >= 3.

Original entry on oeis.org

0, 1, 11, 21, 121, 131, 231, 331, 1331, 1341, 1441, 1541, 2541, 2641, 3641, 4641, 14641, 14651, 14751, 14851, 15851, 15951, 16951, 17951, 27951, 28051, 29051, 30051, 40051, 41051, 51051, 61051, 161051, 161061, 161161, 161261, 162261, 162361, 163361, 164361

Offset: 0

Roger L. Bagula, Mar 15 2006

nonn

From Gary W. Adamson, Aug 30 2016: (Start)
Let M =
    1,  0,  0, 0, 0, ...
   11,  0,  0, 0, 0, ...
   10,  1,  0, 0, 0, ...
    0, 11,  0, 0, 0, ...
    0, 10,  1, 0, 0, ...
    0,  0, 11, 0, 0, ...
    0,  0, 10, 1, 0, ...
  ...
Then lim_{k->infinity} M^k converges to a single nonzero column giving the sequence.
The sequence divided by its aerated variant is (1, 11, 10, 0, 0, 0, ...). (End)

Alois P. Heinz, Table of n, a(n) for n = 0..16383 (first 2501 terms from G. C. Greubel)
H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62, 19-22, 1977.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 27, 33.
Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant

Cf. A000120, A006046, A077465, A084230.

Cf. A130665, A116520, A130667, A116522, A161342, A116526, A360189.

a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 11*a(n/2) else 10*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n),n=0..42);

b[0] := 0; b[1] := 1; b[n_?EvenQ] := b[n] = 11*b[n/2]; b[n_?OddQ] := b[n] = 10*b[(n - 1)/2] + b[(n + 1)/2]; a = Table[b[n], {n, 1, 25}]

Let r(x) = (1 + 11x + 10x^2).  The sequence is r(x) * r(x^2) * r(x^4) * r(x^8) * ... - Gary W. Adamson, Aug 30 2016
a(n) = Sum_{k=0..n-1} 10^wt(k), where wt = A000120. - Mike Warburton, Mar 14 2019
a(n) = Sum_{k=0..floor(log_2(n))} 10^k*A360189(n-1,k). - Alois P. Heinz, Mar 06 2023

Edited by N. J. A. Sloane, Apr 16 2005

A116526 a(0)=1, a(1)=1, a(n) = 9a(n/2) for even n >= 2, and a(n) = 8a((n-1)/2) + a((n+1)/2) for odd n >= 3.

Original entry on oeis.org

0, 1, 9, 17, 81, 89, 153, 217, 729, 737, 801, 865, 1377, 1441, 1953, 2465, 6561, 6569, 6633, 6697, 7209, 7273, 7785, 8297, 12393, 12457, 12969, 13481, 17577, 18089, 22185, 26281, 59049, 59057, 59121, 59185, 59697, 59761, 60273, 60785, 64881, 64945, 65457, 65969

Offset: 0

Roger L. Bagula, Mar 15 2006

nonn

A 9-divide version of A084230.
The interest this one has is in the prime form of even odd 2^n+1, 2^n.
From Gary W. Adamson, Aug 30 2016: (Start)
Let M =
  1, 0, 0, 0, 0, ...
  9, 0, 0, 0, 0, ...
  8, 1, 0, 0, 0, ...
  0, 9, 0, 0, 0, ...
  0, 8, 1, 0, 0, ...
  0, 0, 9, 0, 0, ...
  0, 0, 8, 1, 0, ...
  ...
Then M^k converges to a single nonzero column giving the sequence.
The sequence divided by its aerated variant is (1, 9, 8, 0, 0, 0, ...). (End)

Alois P. Heinz, Table of n, a(n) for n = 0..16383 (first 2501 terms from G. C. Greubel)
H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62, 19-22, 1977.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 27, 33.
Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant

Cf. A000120, A006046, A077465, A130665, A116520, A130667, A116522, A161342, A116525, A360189.

a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 9*a(n/2) else 8*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n),n=0..45);

b[0] := 0; b[1] := 1; b[n_?EvenQ] := b[n] = 9*b[n/2]; b[n_?OddQ] := b[n] = 8*b[(n - 1)/2] + b[(n + 1)/2]; a = Table[b[n], {n, 1, 25}]

a(n) = Sum_{k=0..n-1} 8^wt(k), where wt = A000120. - Mike Warburton, Mar 14 2019

a(n) = Sum_{k=0..floor(log_2(n))} 8^k*A360189(n-1,k). - Alois P. Heinz, Mar 06 2023

Edited by N. J. A. Sloane, Apr 16 2006

cp's OEIS Frontend

A268526 a(n) = r*a(ceiling(n/2))+s*a(floor(n/2)) with a(1)=1 and (r,s)=(3,2).

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A268527 a(n) = r*a(ceiling(n/2))+s*a(floor(n/2)) with a(1)=1 and (r,s)=(4,1).

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A382731 Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 8.

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A116522 a(0)=1, a(1)=1, a(n)=7*a(n/2) for n=2,4,6,..., a(n)=6*a((n-1)/2)+a((n+1)/2) for n=3,5,7,....

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A134659 Total number of odd coefficients in (1+x+x^2)^k for k=0,...,n.

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Mathematica

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A231348 Number of triangles after n-th stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).

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A014370 If n = binomial(b,2) + binomial(c,1), b > c >= 0 then a(n) = binomial(b+1,3) + binomial(c+1,2).

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A074330 a(n) = Sum_{k=1..n} 2^b(k) where b(k) denotes the number of 1's in the binary representation of k.

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A268526 a(n) = ra(ceiling(n/2))+sa(floor(n/2)) with a(1)=1 and (r,s)=(3,2).

A268527 a(n) = ra(ceiling(n/2))+sa(floor(n/2)) with a(1)=1 and (r,s)=(4,1).

A116522 a(0)=1, a(1)=1, a(n)=7a(n/2) for n=2,4,6,..., a(n)=6a((n-1)/2)+a((n+1)/2) for n=3,5,7,....

A116525 a(0)=1, a(1)=1, a(n) = 11a(n/2) for even n, and a(n) = 10a((n-1)/2) + a((n+1)/2) for odd n >= 3.

A116526 a(0)=1, a(1)=1, a(n) = 9a(n/2) for even n >= 2, and a(n) = 8a((n-1)/2) + a((n+1)/2) for odd n >= 3.