A001316 -id:A001316 - cp's OEIS frontend

A382725 Number of entries in the n-th row of Pascal's triangle not divisible by 8.

1, 2, 3, 4, 5, 6, 7, 8, 5, 10, 9, 12, 11, 14, 14, 16, 5, 10, 13, 20, 13, 18, 20, 24, 11, 22, 20, 28, 22, 28, 28, 32, 5, 10, 13, 20, 17, 26, 28, 40, 13, 26, 26, 36, 28, 40, 40, 48, 11, 22, 28, 44, 28, 40, 44, 56, 22, 44, 40, 56, 44, 56, 56, 64, 5, 10, 13, 20, 17, 26, 28, 40, 17, 34, 34, 52, 36, 56, 56, 80, 13

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, A006047, A194459, A382720-A382724.

def A382725(n):
    n1 = n>>1
    n2 = n1>>1
    np = ~n
    n100 = (n2&(~n1)&np).bit_count()
    n110 = (n2&n1&np).bit_count()
    n10 = (n1&np).bit_count()
    return ((n100+1<<3)+(n110<<1)+n10*(n10+3))<>3 # Chai Wah Wu, Aug 10 2025

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

A014421 Odd elements in Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 5, 5, 1, 1, 15, 15, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 1, 1, 9, 9, 1, 1, 45, 45, 1, 1, 11, 55, 165, 165, 55, 11, 1, 1, 495, 495, 1, 1, 13, 715, 1287, 1287, 715, 13, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 15, 105, 455, 1365, 3003, 5005

Offset: 0

Mohammad K. Azarian

The number of terms of the n-row is A001316(n). - Michel Marcus, Jan 11 2016

			Triangle starts:
1;
1, 1;
1, 1;
1, 3, 3, 1;
1, 1;
1, 5, 5, 1;
1, 15, 15, 1;
1, 7, 21, 35, 35, 21, 7, 1;
...

Robert Israel, Table of n, a(n) for n = 0..10070 (rows 0 to 375, flattened)
Arvind Ayyer, Amritanshu Prasad, Steven Spallone, Odd partitions in Young's lattice, arXiv:1601.01776 [math.CO], 2016. See Fig. 3.
A. M. Reiter, Determining the dimension of fractals generated by Pascal’s triangle, Fibonacci Quart, 31(2):112-120, 1993.

Cf. A001316, A007318, A014414.

Cf. A143333. [From Reinhard Zumkeller, Oct 24 2010]

select(type, [seq(seq(binomial(n,k),k=0..n),n=0..20)],odd); # Robert Israel, Jan 11 2016

Select[ Flatten[ Table[ Binomial[ n, i ], {n, 0, 20}, {i, 0, n} ] ], OddQ ]

tabf(nn) = {for (n=0, nn, for (k=0, n, b = binomial(n, k); if (b % 2, print1(b, ", "))); print(););} \\ Michel Marcus, Jan 11 2016

More terms from Erich Friedman
Keyword tabl replaced by tabf by Reinhard Zumkeller, Oct 21 2010
Offset changed to 0 by Michel Marcus, Jan 11 2016

A065040 Triangle read by rows: T(m,k) = exponent of the highest power of 2 dividing the binomial coefficient binomial(m,k).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 2, 3, 1, 3, 2, 3, 0, 0, 0, 2, 2, 1, 1, 2, 2, 0, 0, 0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0

Offset: 0

Claude Lenormand (hlne.lenormand(AT)voono.net), Nov 05 2001

T(m,k) is the number of 'carries' that occur when adding k and m-k in base 2 using the traditional addition algorithm. - Tom Edgar, Jun 10 2014

			Triangle begins:
[0]
[0, 0]
[0, 1, 0]
[0, 0, 0, 0]
[0, 2, 1, 2, 0]
[0, 0, 1, 1, 0, 0]
[0, 1, 0, 2, 0, 1, 0]
[0, 0, 0, 0, 0, 0, 0, 0]
[0, 3, 2, 3, 1, 3, 2, 3, 0]
[0, 0, 2, 2, 1, 1, 2, 2, 0, 0]
[0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0]
... - _N. J. A. Sloane_, Aug 21 2021

Antti Karttunen, Table of n, a(n) for n = 0..10010; the first 141 antidiagonals, flattened
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Row sums: A187059.

Cf. A007318, A007814, A001511, A000120, A047999, A049606, A000680, A048881, A011371, A005187, A000265, A001316, A001317, A243759.

A065040 := (n, k) -> padic[ordp](binomial(n, k), 2):
seq(seq(A065040(n,k), k=0..n), n=0..13); # Peter Luschny, Aug 15 2017

T[m_, k_] := IntegerExponent[Binomial[m, k], 2]; Table[T[m, k], {m, 0, 13}, {k, 0, m}] // Flatten (* Jean-François Alcover, Oct 06 2016 *)

T(m,k)=hammingweight(k)+hammingweight(m-k)-hammingweight(m)
for(m=0,9,for(k=0,m,print1(T(m,k)", "))) \\ Charles R Greathouse IV, Mar 26 2013

As an array f(i,j) = f(j,i) = T(i+j,j) read by antidiagonals: f(0,j) = 0, f(1,j) = A007814(j+1), f(i,j) = Sum_{k=0..i-1} (f(1,j+k) - f(1,k)). [corrected by Kevin Ryde, Oct 07 2021]
The n-th term a(n) is equal to the binomial coefficient binomial(m,k), where m = floor((1+sqrt(8*n+1))/2) - 1 and k = n - m(m+1)/2. Also a(n) = g(m) - g(k) - g(m-k), where g(x) = Sum_{i=1..floor(log_2(x))} floor(x/2^i), m = floor((1+sqrt(8*n+1))/2) - 1, k = n - m(m+1)/2. - Hieronymus Fischer, May 05 2007
T(m,k) <= log_2 m, for m > 0. - Charles R Greathouse IV, Mar 26 2013
T(m,k) = log_2(A082907(m,k)). - Tom Edgar, Jun 10 2014
From Antti Karttunen, Oct 28 2014: (Start)
a(n) = A007814(A007318(n)).
a(n) * A047999(n) = 0 and a(n) + A047999(n) > 0 for all n.
(End)

Name clarified by Antti Karttunen, Oct 28 2014

A089898 Product of (digits of n each incremented by 1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 8, 16, 24, 32, 40, 48, 56, 64

Offset: 0

Marc LeBrun, Nov 13 2003

Sum of products of all subsets of digits of n (with the empty subset contributing 1).
Number of nonnegative values k such that the lunar sum of k and n is n.
First 100 values are 10 X 10 multiplication table, read by rows/columns.

			a(12)=6 since (1+1)*(2+1)=2*3=6 and since (1*2)+(1)+(2)+(1)=2+1+2+1=6 and since the lunar sum of 12 with any of the six values {0,1,2,10,11,12} is 12.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, arXiv:1107.1130 [math.NT], 2011. [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]

Cf. A007954, A087061, A001316, A006047.

a089898 n = if n < 10 then n + 1 else (d + 1) * a089898 n'
            where (n', d) = divMod n 10
-- Reinhard Zumkeller, Jul 06 2014

seq(convert(map(`+`,convert(n,base,10),1),`*`), n = 0 .. 1000); # Robert Israel, Nov 17 2014

a089898[n_Integer] :=
Prepend[Array[Times @@ (IntegerDigits[#] + 1) &, n], 1]; a089898[77] (* Michael De Vlieger, Dec 22 2014 *)

a(n) = my(d=digits(n)); prod(i=1, #d, d[i]+1); \\ Michel Marcus, Apr 06 2014

a(n) = vecprod(apply(x->x+1, digits(n))); \\ Michel Marcus, Feb 01 2023

a(n) = a(floor(n/10))*(1+(n mod 10)). - Robert Israel, Nov 17 2014

G.f. g(x) satisfies g(x) = (10*x^11 - 11*x^10 + 1)*g(x^10)/(x-1)^2. - Robert Israel, Nov 17 2014

A100307 Modulo 2 binomial transform of 3^n.

Original entry on oeis.org

1, 4, 10, 40, 82, 328, 820, 3280, 6562, 26248, 65620, 262480, 538084, 2152336, 5380840, 21523360, 43046722, 172186888, 430467220, 1721868880, 3529831204, 14119324816, 35298312040, 141193248160, 282472589764, 1129890359056

Offset: 0

Paul Barry, Dec 06 2004

3^n may be retrieved through 3^n = Sum_{k=0..n} (-1)^A010060(n-k)*(binomial(n,k) mod 2)*a(k).

Gheorghe Coserea, Table of n, a(n) for n = 0..200
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Cf. A001316, A001317, A038183, A100308, A100309, A100310, A100311.

[(&+[3^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 03 2023

Table[Sum[Mod[Binomial[n,k],2]3^k,{k,0,n}],{n,0,40}] (* Harvey P. Dale, Aug 28 2013 *)

a(n) = subst(lift((Mod(1,2)+'x)^n), 'x, 3); \\ Gheorghe Coserea, Jun 11 2016

def A100307(n): return sum((bool(~n&n-k)^1)*3**k for k in range(n+1)) # Chai Wah Wu, May 02 2023

[sum((binomial(n,k)%2)*3^k for k in [0..n]) for n in [0..50]] # Tom Edgar, Oct 11 2015

a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*3^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(3^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(3^(2^k)+1)^r), where r > 0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(3^(2^k)+1)) = 1.391980...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 2/3.
a(2^n) = 3^(2^n)+1, n >= 0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:
a(2^t*n+2^(t-1)) = 8*(3^(2^(t-1)+1))/(3^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2.
In particular, for t=2,3,4, we have the following formulas:
  a(4*n+2)  =         10 * a(4*n),
  a(8*n+4)  =     (41/5) * a(8*n+2),
  a(16*n+8) = (3281/410) * a(16*n+6), etc. (End)
From Tom Edgar, Oct 11 2015: (Start)
a(n) = Product_{b_j != 0} a(2^j) where n = Sum_{j>=0} b_j*2^j is the binary representation of n.
a(2*k+1) = 4*a(2*k). (End)

A100308 Modulo 2 binomial transform of 5^n.

Original entry on oeis.org

1, 6, 26, 156, 626, 3756, 16276, 97656, 390626, 2343756, 10156276, 60937656, 244531876, 1467191256, 6357828776, 38146972656, 152587890626, 915527343756, 3967285156276, 23803710937656, 95520019531876, 573120117191256

Offset: 0

Paul Barry, Dec 06 2004

5^n may be retrieved through 5^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2)*a(k).

Robert Israel, Table of n, a(n) for n = 0..1429
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Cf. A001316, A001317, A038183, A100307, A100309, A100310, A100311.

[(&+[5^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 03 2023

f:= proc(n) local L,M;
   L:= convert(n,base,2);
   mul(1+5^(2^(k-1)), k = select(t -> L[t]=1, [$1..nops(L)]));
end proc:
map(f, [$0..30]); # Robert Israel, Aug 26 2018

a[n_]:= Sum[Mod[Binomial[n, k], 2] 5^k, {k, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Sep 19 2018 *)

def A100308(n): return sum((bool(~n&n-k)^1)*5**k for k in range(n+1)) # Chai Wah Wu, May 02 2023

def A100308(n): return sum(5^k*(binomial(n, k)%2) for k in range(n+1))
[A100308(n) for n in range(41)] # G. C. Greubel, Feb 03 2023

a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*5^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(5^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(5^(2^k)+1)^r), where r > 0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(5^(2^k)+1)) = 1.2134769...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.8.
a(2^n) = 5^(2^n) + 1, n >= 0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:
a(2^t*n+2^(t-1)) = 24*(5^(2^(t-1)+1))/(5^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2.
In particular, for t=2,3,4, we have the following formulas:
  a(4*n+2)  =            26 * a(4*n),
  a(8*n+4)  =      (313/13) * a(8*n+2),
  a(16*n+8) = (195313/8138) * a(16*n+6), etc. (End)

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

A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

Original entry on oeis.org

1, 1, 2, 1, 17, 26, 2, 62, 192, 60, 1, 1382, 7192, 5097, 502, 2, 21844, 171511, 217186, 55196, 2036, 2, 929569, 10262046, 20376780, 9893440, 1089330, 16356, 4, 6404582, 94582204, 271154544, 215114420, 48673180, 2567568, 16376, 1

Offset: 1

Johannes W. Meijer, May 24 2009

In A160464 we defined the ES1 matrix by ES1[2*m-1,n=1] and in A094665 it was shown that the n-th term of the coefficients of matrix row ES1[1-2*m,n] for m >= 1 can be generated with the RES1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRES1(z,1-2*m) = sum(RES1(1-2*m,n)*z^(n-1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,1-2*m) = (-1)*RE(z,1-2*m)/(2*p(m-1)*(z-1)^(m)). The p(m-1), m >= 1, sequence is Gould's sequence A001316.
The coefficients of the RE(z,1-2*m) polynomials lead to the triangle given above.
The E(z,n) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
Some of our results are conjectures based on numerical evidence.

			The first few rows are:
[1]
[1]
[2, 1]
[17, 26, 2]
[62, 192, 60, 1]
The first few polynomials RE(z,m) are:
RE(z,-1) = 1
RE(z,-3) = 1
RE(z,-5) = 2+z
RE(z,-7) = 17+26*z+2*z^2
The first few GFRES1(z,m) are:
GFRES1(z,-1) = -(1/1)*(1)/(2*(z-1)^1)
GFRES1(z,-3) = -(1/2)*(1)/(2*(z-1)^2)
GFRES1(z,-5) = -(1/2)*(2+z)/(2*(z-1)^3)
GFRES1(z,-7) = -(1/4)*(17+26*z+2*z^2)/(2*(z-1)^4)

Grzegorz Rzadkowski, M Urlinska, A Generalization of the Eulerian Numbers, arXiv preprint arXiv:1612.06635, 2016

Cf. A160464, A094665 and A083061.
For the Eulerian numbers E(n, k) see A008292.
The p(n) sequence equals Gould's sequence A001316.
The first right hand column of the triangle equals A048896.
The first left hand column equals A160469.
The row sums equal the absolute values of A117972.

nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i-1, x), x): for na from 0 to dgr do c(na) := coeff(T(i-1, x), x, na) od: T(i-1, x+1) := 0: for nb from 0 to dgr do T(i-1, x+1) := T(i-1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i-1, nc+1) := coeff(T(i-1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i-1, x+1) - 2*x^2*T(i-1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((-1)^(m-q)*(q^k)*binomial(k+1, m-q), q=1..m) od: fx(p) := (-1)^(p+1) * (sum(E(r, k)*z^(k-r), r=1..k))/(z-1)^(p+1): GF(-(2*p+1)) := sort(simplify(((-1)^p* 1/2^(p+1)) * sum(ECGP(k-1, k-s)*fx(k-s), s=0..k-1)), ascending): NUMGF(-(2*p+1)) := -numer(GF(-(2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF(-(2*p+1)), z, n-1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n-1)/2+m+1) := A(n+1, m) end do end do: a := n-> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);

T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m - 1}, x Sum[ a[k] a[n - k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n - 1}]] (* Michael Somos, Apr 22 2020 *)

Edited by Johannes W. Meijer, Sep 23 2012

A160721 First differences of A160720.

Original entry on oeis.org

1, 4, 4, 12, 4, 12, 12, 28, 4, 12, 12, 28, 12, 28, 28, 60, 4, 12, 12, 28, 12, 28, 28, 60, 12, 28, 28, 60, 28, 60, 60, 124, 4, 12, 12, 28, 12, 28, 28, 60, 12, 28, 28, 60, 28, 60, 60, 124, 12, 28, 28, 60, 28, 60, 60, 124, 28, 60, 60, 124, 60, 124, 124, 252, 4, 12, 12, 28, 12, 28, 28

Offset: 1

Omar E. Pol, May 25 2009, May 29 2009

This sequence is related to the Sierpinski triangle and to Gould's sequence A001316. - Omar E. Pol, Jul 23 2009

When written as a irregular triangle in which row lengths are A011782 it appears that right border gives A173033. - Omar E. Pol, Mar 20 2013

			From _Omar E. Pol_, Mar 20 2013 (Start):
Triangle begins:
1;
4;
4,12;
4,12,12,28;
4,12,12,28,12,28,28,60;
4,12,12,28,12,28,28,60,12,28,28,60,28,60,60,124;
4,12,12,28,12,28,28,60,12,28,28,60,28,60,60,124,12,28,28,60,28,60,60,124,28,60,60,124,60,124,124,252;
(End)

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.]
David Applegate, The movie version
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to toothpick sequences

Cf. A000120, A001316, A038573, A139250, A139251, A160720, A160722, A160723.

a(1)=1. Observation: It appears that a(n) = 4*A038573(n-1), n>1. [From Omar E. Pol, Jul 23 2009]. This formula is correct! - N. J. A. Sloane, Jan 23 2016

More terms from R. J. Mathar, Jul 14 2009

cp's OEIS Frontend

A382725 Number of entries in the n-th row of Pascal's triangle not divisible by 8.

Views

Author

Keywords

Links

Crossrefs

Programs

Python

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

Views

Author

Keywords

Links

Crossrefs

Programs

Python

A014421 Odd elements in Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Extensions

A065040 Triangle read by rows: T(m,k) = exponent of the highest power of 2 dividing the binomial coefficient binomial(m,k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A089898 Product of (digits of n each incremented by 1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

A100307 Modulo 2 binomial transform of 3^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Sage

Formula

A100308 Modulo 2 binomial transform of 5^n.

Views

Author

Keywords

Comments

Links

Crossrefs