A191372 -id:A191372 - cp's OEIS frontend

A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2n) = a(n), a(2n+1) = a(n) + a(n+1).

0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6, 1, 7, 6, 11, 5, 14, 9, 13, 4, 15, 11, 18, 7, 17, 10, 13, 3, 14, 11, 19, 8, 21, 13, 18, 5, 17, 12, 19

Offset: 0

N. J. A. Sloane

Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
  column 0 1 2 3 4 5 6 7 8 9 ...
  row 0: 0
  row 1: 1
  row 2: 1,2
  row 3: 1,3,2,3
  row 4: 1,4,3,5,2,5,3,4
  row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
  row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
  ...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
From N. J. A. Sloane, Oct 15 2017: (Start)
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
     1 (n >= 1),
     n (n >= 2),
   n-1 (n >= 3),
  2n-3 (n >= 3),
   n-2 (n >= 4),
  3n-7 (n >= 4),
  ...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise.
(End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006
a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1].
a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014
a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006
The coefficients of the inverse of the g.f. of this sequence form A073469 and are related to binary partitions A000123. - Philippe Flajolet, Sep 06 2008
It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
   1;
   1, 0;
   1, 1, 0;
   0, 1, 0, 0;
   0, 1, 1, 0, 0;
   0, 0, 1, 0, 0, 0;
   0, 0, 1, 1, 0, 0, 0;
   ...
   Then this sequence A002487 (without initial 0) is the first column of lim_{n->oo} M^n. (Cf. A026741.) - Gary W. Adamson, Dec 11 2009 [Edited by M. F. Hasler, Feb 12 2017]
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010
Equals row 1 in an infinite array shown in A178568, sequences of the form
a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010
Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011
The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011
Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012
Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013
Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function (A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014
Back in May 1995, it was proved that A000360 is the modulo 3 mapping, (+1,-1,+0)/2, of this sequence A002487 (without initial 0). - M. Jeremie Lafitte (Levitas), Apr 24 2017
Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017
The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019
Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1).  For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7. - Gary W. Adamson, Dec 18 2023

			Stern's diatomic array begins:
  1,1,
  1,2,1,
  1,3,2,3,1,
  1,4,3,5,2,5,3,4,1,
  1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
  1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
  ...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0;  L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - _Yosu Yurramendi_, Jul 13 2016
From _I. V. Serov_, Jun 01 2017: (Start)
a(n) is the length of the Christoffel word chf(n):
n  chf(n) A070939(n)   a(n)
1   '-'       1          1
2   '+'       2          1
3   '+-'      2          2
4   '-'       3          1
5   '--+'     3          3
6   '-+'      3          2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - _Michael Somos_, Jun 25 2019

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Index entries for "core" sequences
Index entries for fraction trees
Index entries for sequences related to enumerating the rationals
Index entries for sequences related to Stern's sequences
Index entries for sequences related to trees

Cf. A000120, A000123, A000360, A001045, A002083, A011655, A020950, A026741, A037227, A046815, A070871, A070872, A071883, A073459, A084091, A101624, A126606, A168081, A174980, A174981, A178239, A178568, A212288, A213369, A260443, A277020, A277325, A287729, A287730, A293160.
Record values are in A212289.
If the 1's are replaced by pairs of 1's we obtain A049456.
Inverse: A020946.
Cf. a(A001045(n)) = A000045(n). a(A062092(n)) = A000032(n+1).
Cf. A064881-A064886 (Stern-Brocot subtrees).
A column of A072170.
Cf. A049455 for the 0,1 version of Stern's diatomic array.
Cf. A000119, A262097 for analogous sequences in other bases and A277189, A277315, A277328 for related sequences with similar graphs.
Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.

a002487 n = a002487_list !! n
a002487_list = 0 : 1 : stern [1] where
   stern fuscs = fuscs' ++ stern fuscs' where
     fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
   interleave []     ys = ys
   interleave (x:xs) ys = x : interleave ys xs
-- Reinhard Zumkeller, Aug 23 2011

using Nemo
function A002487List(len)
    a, A = QQ(0), [0,1]
    for n in 1:len
        a = next_calkin_wilf(a)
        push!(A, denominator(a))
    end
A end
A002487List(91) |> println # Peter Luschny, Mar 13 2018

[&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019

A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq(A002487(n),n=0..91);
A002487 := proc(m) local a,b,n; a := 1; b := 0; n := m; while n>0 do if type(n,odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq(A002487(n),n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k,n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq(A002487(n),n=0..91); # Thomas Wieder, Jan 13 2008
# next Maple program:
a:= proc(n) option remember; `if`(n<2, n,
      (q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Feb 11 2021
fusc := proc(n) local a, b, c; a := 1; b := 0;
    for c in convert(n, base, 2) do
        if c = 0 then a := a + b else b := a + b fi od;
    b end:
seq(fusc(n), n = 0..91); # Peter Luschny, Nov 09 2022
Stern := proc(n, u) local k, j, b;
    b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
    ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024

a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
NestList[Onemore, {1}, 5] // Flatten  (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]]  (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
a[0] = 0; a[1] = 1;
Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *)
nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)

{a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};

fusc(n)=local(a=1,b=0);while(n>0,if(bitand(n,1),b+=a,a+=b);n>>=1);b \\ Charles R Greathouse IV, Oct 05 2008

A002487(n,a=1,b=0)=for(i=0,logint(n,2),if(bittest(n,i),b+=a,a+=b));b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019

from functools import lru_cache
@lru_cache(maxsize=None)
def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021

def a(n):
    a, b = 1, 0
    while n > 0:
        if n & 1:
            b += a
        else:
            a += b
        n >>= 1
    return b
# Reza K Ghazi, Dec 29 2021

def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022

# (fast way for big vectors)
from math import log, ceil
import numpy
how_many_terms = 2**20  # (Powers of 2 recommended but other integers are also possible.)
A002487, A002487[1]  = numpy.zeros(2**(ce:=ceil(log(how_many_terms,2))), dtype=object), 1
for exponent in range(1,ce):
    L, L2 = 2**exponent, 2**(exponent+1)
    A002487[L2 - 1] = exponent + 1
    A002487[L:L2][::2] = A002487[L >> 1: L]
    A002487[L + 1:L2 - 2][::2] = A002487[L:L2 - 3][::2]  +  A002487[L + 2:L2 - 1][::2]
print(list(A002487[0:100])) # Karl-Heinz Hofmann, Jul 22 2025

N <- 50 # arbitrary
a <- 1
for (n in 1:N)
{
  a[2*n    ] = a[n]
  a[2*n + 1] = a[n] + a[n+1]
  a
}
a
# Yosu Yurramendi, Oct 04 2014

# Given n, compute a(n) by taking into account the binary representation of n
a <- function(n){
  b <- as.numeric(intToBits(n))
  l <- sum(b)
  m <- which(b == 1)-1
  d <- 1
  if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
  f <- c(0,1)
  if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
  return(f[l+1])
} # Yosu Yurramendi, Dec 13 2016

# computes the sequence as a vector A, rather than function a() as above.
A <- c(1,1)
maxlevel <- 5 # by choice
for(m in 1:maxlevel) {
  A[2^(m+1)] <- 1
  for(k in 1:(2^m-1)) {
    r <- m - floor(log2(k)) - 1
    A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
}}
A # Yosu Yurramendi, May 08 2018

def A002487(n):
    M = [1, 0]
    for b in n.bits():
        M[b] = M[0] + M[1]
    return M[1]
print([A002487(n) for n in (0..91)])
# For a dual see A174980. Peter Luschny, Nov 28 2017

;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
(definec (A002487 n) (cond ((<= n 1) n) ((even? n) (A002487 (/ n 2))) (else (+ (A002487 (/ (- n 1) 2)) (A002487 (/ (+ n 1) 2))))))
;; Antti Karttunen, Nov 05 2016

a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001
Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004.  Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006
A079978(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
a(n) = A126606(n + 1) / 2. - Reikku Kulon, Oct 05 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009
a(2^(p+2)*n+2^(p+1)-1) - a(2^(p+1)*n+2^p-1) = A007306(n+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 07 2013
a(2*n-1) = A007306(n), n > 0. - Yosu Yurramendi, Jun 23 2014
a(n*2^m) = a(n), m>0, n > 0. - Yosu Yurramendi, Jul 03 2014
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016
From Yosu Yurramendi, Jul 13 2016: (Start)
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(n) = A001222(A260443(n)) = A000120(A277020(n)). Also a(n) = A000120(A101624(n-1)) for n >= 1. - Antti Karttunen, Nov 05 2016
(a(n-1) + a(n+1))/a(n) = A037227(n) for n >= 1. - Peter Bala, Feb 07 2017
a(0) = 0; a(3n) = 2*A000360(3n-1); a(3n+1) = 2*A000360(3n) - 1; a(3n+2) = 2*A000360(3n+1) + 1. - M. Jeremie Lafitte (Levitas), Apr 24 2017
From I. V. Serov, Jun 14 2017: (Start)
a(n) = A287896(n-1) - 1*A288002(n-1) for n > 1;
a(n) = A007306(n-1) - 2*A288002(n-1) for n > 1. (End)
From Yosu Yurramendi, Feb 14 2018: (Start)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1)       + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m,  a(2^m) = 1, a(0) = 0. (End)
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
a(n) = A000120(A168081(n)). - Karl-Heinz Hofmann, Jun 16 2025

Additional references and comments from Len Smiley, Joshua Zucker, Rick L. Shepherd and Herbert S. Wilf
Typo in definition corrected by Reinhard Zumkeller, Aug 23 2011
Incorrect formula deleted and text edited by Johannes W. Meijer, Feb 07 2013

A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1

Offset: 0

N. J. A. Sloane

Restored the alternative spelling of Sierpinski to facilitate searching for this triangle using regular-expression matching commands in ASCII. - N. J. A. Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by "Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2. - Philippe Deléham, Oct 02 2003
Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche and Berthe]
Also triangle formed by reading triangles A011117, A028338, A039757, A059438, A085881, A086646, A086872, A087903, A104219 mod 2. - Philippe Deléham, Jun 18 2005
J. H. Conway writes (in Math Forum): at least the first 31 rows give odd-sided constructible polygons (sides 1, 3, 5, 15, 17, ... see A001317). The 1's form a Sierpiński sieve. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
When regarded as an infinite lower triangular matrix, its inverse is a (-1,0,1)-matrix with zeros undisturbed and the nonzero entries in every column form the Prouhet-Thue-Morse sequence (1,-1,-1,1,-1,1,1,-1,...) A010060 (up to relabeling). - David Callan, Oct 27 2006
Triangle read by rows: antidiagonals of an array formed by successive iterates of running sums mod 2, beginning with (1, 1, 1, ...). - Gary W. Adamson, Jul 10 2008
T(n,k) = A057427(A143333(n,k)). - Reinhard Zumkeller, Oct 24 2010
The triangle sums, see A180662 for their definitions, link Sierpiński’s triangle A047999 with seven sequences, see the crossrefs. The Kn1y(n) and Kn2y(n), y >= 1, triangle sums lead to the Sierpiński-Stern triangle A191372. - Johannes W. Meijer, Jun 05 2011
Used to compute the total Steifel-Whitney cohomology class of the Real Projective space. This was an essential component of the proof that there are no product operations without zero divisors on R^n for n not equal to 1, 2, 4 or 8 (real numbers, complex numbers, quaternions, Cayley numbers), proved by Bott and Milnor. - Marcus Jaiclin, Feb 07 2012
T(n,k) = A134636(n,k) mod 2. - Reinhard Zumkeller, Nov 23 2012
T(n,k) = 1 - A219463(n,k), 0 <= k <= n. - Reinhard Zumkeller, Nov 30 2012
From Vladimir Shevelev, Dec 31 2013: (Start)
Also table of coefficients of polynomials s_n(x) of degree n which are defined by formula s_n(x) = Sum_{i=0..n} (binomial(n,i) mod 2)*x^k. These polynomials we naturally call Sierpiński's polynomials. They also are defined by the recursion: s_0(x)=1, s_(2*n+1)(x) = (x+1)*s_n(x^2), n>=0, and s_(2*n)(x) = s_n(x^2), n>=1.
Note that: s_n(1)  = A001316(n),
           s_n(2)  = A001317(n),
           s_n(3)  = A100307(n),
           s_n(4)  = A001317(2*n),
           s_n(5)  = A100308(n),
           s_n(6)  = A100309(n),
           s_n(7)  = A100310(n),
           s_n(8)  = A100311(n),
           s_n(9)  = A100307(2*n),
           s_n(10) = A006943(n),
           s_n(16) = A001317(4*n),
           s_n(25) = A100308(2*n), etc.
The equality s_n(10) = A006943(n) means that sequence A047999 is obtained from A006943 by the separation by commas of the digits of its terms. (End)
Comment from N. J. A. Sloane, Jan 18 2016: (Start)
Take a diamond-shaped region with edge length n from the top of the triangle, and rotate it by 45 degrees to get a square S_n. Here is S_6:
  [1, 1, 1, 1, 1, 1]
  [1, 0, 1, 0, 1, 0]
  [1, 1, 0, 0, 1, 1]
  [1, 0, 0, 0, 1, 0]
  [1, 1, 1, 1, 0, 0]
  [1, 0, 1, 0, 0, 0].
Then (i) S_n contains no square (parallel to the axes) with all four corners equal to 1 (cf. A227133); (ii) S_n can be constructed by using the greedy algorithm with the constraint that there is no square with that property; and (iii) S_n contains A064194(n) 1's. Thus A064194(n) is a lower bound on A227133(n). (End)
See A123098 for a multiplicative encoding of the rows, i.e., product of the primes selected by nonzero terms; e.g., 1 0 1 => 2^1 * 3^0 * 5^1. - M. F. Hasler, Sep 18 2016
From Valentin Bakoev, Jul 11 2020: (Start)
The Sierpinski's triangle with 2^n rows is a part of a lower triangular matrix M_n of dimension 2^n X 2^n. M_n is a block matrix defined recursively: M_1= [1, 0], [1, 1], and for n>1, M_n = [M_(n-1), O_(n-1)], [M_(n-1), M_(n-1)], where M_(n-1) is a block matrix of the same type, but of dimension 2^(n-1) X 2^(n-1), and O_(n-1) is the zero matrix of dimension 2^(n-1) X 2^(n-1). Here is how M_1, M_2 and M_3 look like:
1 0     1 0 0 0      1 0 0 0 0 0 0 0
1 1     1 1 0 0      1 1 0 0 0 0 0 0   - It is seen the self-similarity of the
        1 0 1 0      1 0 1 0 0 0 0 0     matrices M_1, M_2, ..., M_n, ...,
        1 1 1 1      1 1 1 1 0 0 0 0     analogously to the Sierpinski's fractal.
                     1 0 0 0 1 0 0 0
                     1 1 0 0 1 1 0 0
                     1 0 1 0 1 0 1 0
                     1 1 1 1 1 1 1 1
M_n can also be defined as M_n = M_1 X M_(n-1) where X denotes the Kronecker product. M_n is an important matrix in coding theory, cryptography, Boolean algebra, monotone Boolean functions, etc. It is a transformation matrix used in computing the algebraic normal form of Boolean functions. Some properties and links concerning M_n can be seen in LINKS. (End)
Sierpinski's gasket has fractal (Hausdorff) dimension log(A000217(2))/log(2) = log(3)/log(2) = 1.58496... (and cf. A020857). This gasket is the first of a family of gaskets formed by taking the Pascal triangle (A007318) mod j, j >= 2 (see CROSSREFS). For prime j, the dimension of the gasket is log(A000217(j))/log(j) = log(j(j + 1)/2)/log(j) (see Reiter and Bondarenko references). - Richard L. Ollerton, Dec 14 2021

			Triangle begins:
              1,
             1,1,
            1,0,1,
           1,1,1,1,
          1,0,0,0,1,
         1,1,0,0,1,1,
        1,0,1,0,1,0,1,
       1,1,1,1,1,1,1,1,
      1,0,0,0,0,0,0,0,1,
     1,1,0,0,0,0,0,0,1,1,
    1,0,1,0,0,0,0,0,1,0,1,
   1,1,1,1,0,0,0,0,1,1,1,1,
  1,0,0,0,1,0,0,0,1,0,0,0,1,
  ...

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004).
John W. Milnor and James D. Stasheff, Characteristic Classes, Princeton University Press, 1974, pp. 43-49 (sequence appears on p. 46).
H.-O. Peitgen, H. Juergens and D. Saupe: Chaos and Fractals (Springer-Verlag 1992), p. 408.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

N. J. A. Sloane, Table of n, a(n) for n = 0..10584 [First 144 rows, flattened; first 50 rows from T. D. Noe].
J.-P. Allouche and V. Berthe, Triangle de Pascal, complexité et automates, Bulletin of the Belgian Mathematical Society Simon Stevin 4.1 (1997): 1-24.
J.-P. Allouche, F. v. Haeseler, H.-O. Peitgen and G. Skordev, Linear cellular automata, finite automata and Pascal's triangle, Discrete Appl. Math. 66 (1996), 1-22.
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.],
J. Baer, Explore patterns in Pascal's Triangle
Valentin Bakoev, Fast Bitwise Implementation of the Algebraic Normal Form Transform, Serdica J. of Computing 11 (2017), No 1, 45-57.
Valentin Bakoev, Properties and links concerning M_n
Thomas Baruchel, Flattening Karatsuba's Recursion Tree into a Single Summation, SN Computer Science (2020) Vol. 1, Article No. 48.
Thomas Baruchel, A non-symmetric divide-and-conquer recursive formula for the convolution of polynomials and power series, arXiv:1912.00452 [math.NT], 2019.
A. Bogomolny, Dot Patterns and Sierpinski Gasket
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
Paul Bradley and Peter Rowley, Orbits on k-subsets of 2-transitive Simple Lie-type Groups, 2014.
E. Burlachenko, Fractal generalized Pascal matrices, arXiv:1612.00970 [math.NT], 2016. See p. 9.
S. Butkevich, Pascal Triangle Applet
David Callan, Sierpinski's triangle and the Prouhet-Thue-Morse word, arXiv:math/0610932 [math.CO], 2006.
B. Cherowitzo, Pascal's Triangle using Clock Arithmetic, Part I
B. Cherowitzo, Pascal's Triangle using Clock Arithmetic, Part II
C. Cobeli, A. Zaharescu, A game with divisors and absolute differences of exponents, arXiv:1411.1334 [math.NT], 2014; Journal of Difference Equations and Applications, Vol. 20, #11, 2014.
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712.
Brady Haran, Chaos Game, Numberphile video, YouTube (April 27, 2017).
I. Kobayashi et al., Pascal's Triangle
Dr. Math, Regular polygon formulas [Broken link?]
Y. Moshe, The distribution of elements in automatic double sequences, Discr. Math., 297 (2005), 91-103.
National Curve Bank, Sierpinski Triangles
Hieu D. Nguyen, A Digital Binomial Theorem, arXiv:1412.3181 [math.NT], 2014.
S. Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
F. Richman, Javascript for computing Pascal's triangle modulo n. Go to this page, then under "Modern Algebra and Other Things", click "Pascal's triangle modulo n".
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29. Also arXiv:1011.6083, 2010.
N. J. A. Sloane, Illustration of rows 0 to 32 (encoignure style)
N. J. A. Sloane, Illustration of rows 0 to 64 (encoignure style)
N. J. A. Sloane, Illustration of rows 0 to 128 (encoignure style)
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Eric Weisstein's World of Mathematics, Sierpiński Sieve, Rule 60, Rule 102
Index entries for sequences related to cellular automata
Index entries for triangles and arrays related to Pascal's triangle
Index entries for sequences generated by sieves

Sequences based on the triangles formed by reading Pascal's triangle mod m: (this sequence) (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
Cf. A007318, A054431, A001317, A008292, A083093, A034931, A034930, A008975, A034932, A166360, A249133, A064194, A227133.
From Johannes W. Meijer, Jun 05 2011: (Start)
A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4), A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3, Ze4). (End)

import Data.Bits (xor)
a047999 :: Int -> Int -> Int
a047999 n k = a047999_tabl !! n !! k
a047999_row n = a047999_tabl !! n
a047999_tabl = iterate (\row -> zipWith xor ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Dec 11 2011, Oct 24 2010

A047999:= func< n,k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
[A047999(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 03 2024

# Maple code for first M rows (here M=10) - N. J. A. Sloane, Feb 03 2016
ST:=[1,1,1]; a:=1; b:=2; M:=10;
for n from 2 to M do ST:=[op(ST),1];
for i from a to b-1 do ST:=[op(ST), (ST[i+1]+ST[i+2]) mod 2 ]; od:
ST:=[op(ST),1];
a:=a+n; b:=a+n; od:
ST; # N. J. A. Sloane
# alternative
A047999 := proc(n,k)
    modp(binomial(n,k),2) ;
end proc:
seq(seq(A047999(n,k),k=0..n),n=0..12) ; # R. J. Mathar, May 06 2016

Mod[ Flatten[ NestList[ Prepend[ #, 0] + Append[ #, 0] &, {1}, 13]], 2] (* Robert G. Wilson v, May 26 2004 *)
rows = 14; ca = CellularAutomaton[60, {{1}, 0}, rows-1]; Flatten[ Table[ca[[k, 1 ;; k]], {k, 1, rows}]] (* Jean-François Alcover, May 24 2012 *)
Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 26 2019 *)
A047999[n_,k_]:= Boole[BitAnd[n-k,k]==0];
Table[A047999[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 03 2025 *)

\\ Recurrence for Pascal's triangle mod p, here p = 2.
p = 2; s=13; T=matrix(s,s); T[1,1]=1;
for(n=2,s, T[n,1]=1; for(k=2,n, T[n,k] = (T[n-1,k-1] + T[n-1,k])%p ));
for(n=1,s,for(k=1,n,print1(T[n,k],", "))) \\ Gerald McGarvey, Oct 10 2009

A011371(n)=my(s);while(n>>=1,s+=n);s
T(n,k)=A011371(n)==A011371(k)+A011371(n-k) \\ Charles R Greathouse IV, Aug 09 2013

T(n,k)=bitand(n-k,k)==0 \\ Charles R Greathouse IV, Aug 11 2016

def A047999_T(n,k):
    return int(not ~n & k) # Chai Wah Wu, Feb 09 2016

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary expansion of k are a subset of the 1's in the binary expansion of n; or equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0 < k < n; T(n,0) = T(n,n) = 1. - Reinhard Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0 < k < n; T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013: (Start)
For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = Product_{i=1..A000120(n)} (x^(2^k_i) + 1),
if the binary expansion of n is n = Sum_{i=1..A000120(n)} 2^k_i;
G.f. Sum_{n>=0} s_n(x)*z^n = Product_{k>=0} (1 + (x^(2^k)+1)*z^(2^k)) (0Let x>1, t>0 be real numbers. Then
Sum_{n>=0} 1/s_n(x)^t = Product_{k>=0} (1 + 1/(x^(2^k)+1)^t);
Sum_{n>=0} (-1)^A000120(n)/s_n(x)^t = Product_{k>=0} (1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
Sum_{n>=0} (-1)^A000120(n)/s_n(x) = 1 - 1/x. (End)
From Valentin Bakoev, Jul 11 2020: (Start)
(See my comment about the matrix M_n.) Denote by T(i,j) the number in the i-th row and j-th column of M_n (0 <= i, j < 2^n). When i>=j, T(i,j) is the j-th number in the i-th row of the Sierpinski's triangle. For given i and j, we denote by k the largest integer of the type k=2^m and kT(i,0) = T(i,i) = 1, or
T(i,j) = 0 if i < j, or
T(i,j) = T(i-k,j), if j < k, or
T(i,j) = T(i-k,j-k), if j >= k.
Thus, for given i and j, T(i,j) can be computed in O(log_2(i)) steps. (End)

Additional links from Lekraj Beedassy, Jan 22 2004

A191488 A companion to Gould’s sequence A001316.

Original entry on oeis.org

4, 6, 8, 10, 8, 12, 16, 18, 8, 12, 16, 20, 16, 24, 32, 34, 8, 12, 16, 20, 16, 24, 32, 36, 16, 24, 32, 40, 32, 48, 64, 66, 8, 12, 16, 20, 16, 24, 32, 36, 16, 24, 32, 40, 32, 48, 64, 68, 16, 24, 32, 40, 32, 48, 64, 72, 32, 48, 64, 80, 64, 96, 128

Offset: 0

Johannes W. Meijer, Jun 05 2011

The row sums of the Sierpinski-Stern triangle A191372 are given by sequence A191487.
The differences diff1(n) = A191487(2*n+3) - A191487(2*n+1) lead to a peculiar number triangle, see the examples. The leading terms of the rows of the diff1(n) triangle clearly stand out from the rest of the terms and are given by A001550(p+1), p>=1; for p=0 this term is 7.
If we ignore the first term of the diff1(n) rows and reverse the order of the remaining terms we get sequence A191488, see the examples; more terms require a higher row number.
Both the diff1(n) and the diff2(n) sequences are related to Gould’s sequence A001316. We ignore the first term and reverse the order of the rest of the terms. The diff2(n) sequence leads directly to A001316, see A191487, while the diff1(n) sequence leads to A001316 in a slightly more complex way. We observe that for Gould’s sequence equation A001316((2*n+1)*2^p-1) = C(p)*A001316(n) with C(p) = 2^p holds, while for its companion A191488 equation A191488((2*n+1)*2^p-1) = C(p)*A001316(n) with C(p) = 2^(p+1)+2 holds; see the Maple program. Furthermore for both sequences a(2^p - 1) = C(p).

			The first few rows of diff1(n) as a triangle, row lengths A000079(p) with p>=0, are:
[7]
[14, 4]
[36, 8, 6, 4]
[98, 16, 12, 8, 10, 8, 6, 4]
[276, 32, 24, 16, 20, 16, 12, 8, 18, 16, 12, 8, 10, 8, 6, 4]
[794, 64, 48, 32, 40, 32, 24, 16, 36, 32, 24, 16, 20, 16, 12, 8, 34, 32, 24, 16, 20, 16, 12, 8, 18, 16, 12, 8, 10, 8, 6, 4]
The first few rows of diff1(n) reversed minus the first term are:
[4]
[4, 6, 8]
[4, 6, 8, 10, 8, 12, 16]
[4, 6, 8, 10, 8, 12, 16, 18, 8, 12, 16, 20, 16, 24, 32]
[4, 6, 8, 10, 8, 12, 16, 18, 8, 12, 16, 20, 16, 24, 32, 34, 8, 12, 16, 20, 16, 24, 32, 36, 16, 24, 32, 40, 32, 48, 64]

Cf. A001316, A191372, A191487.

nmax:=2^6; pmax:=ceil(log(nmax)/log(2)); A001316 := n -> if n<=-1 then 0 else 2^add(i, i=convert(n, base, 2)) fi: C := proc(p): C(p) := 2^(p+1)+2 end: for p from 0 to pmax do for n from 0 to nmax do a((2*n+1)*2^p-1):= C(p)*A001316(n) od: od: seq(a(n), n=0..nmax-2);

a((2*n+1)*2^p - 1) = C(p) * A001316(n) with C(p) = (2^(p+1)+2), p>=0.

a(2^p - 1) = 2^(p+1)+2 = A052548(p+1), p>=0.

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A191487 The row sums of the Sierpinski-Stern triangle A191372.

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A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

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A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

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A191488 A companion to Gould’s sequence A001316.

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A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2n) = a(n), a(2n+1) = a(n) + a(n+1).