A052955 -id:A052955 - cp's OEIS frontend

A163978 a(n) = 2*a(n-2) for n > 2; a(1) = 3, a(2) = 4.

3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152, 3145728

Offset: 1

Klaus Brockhaus, Aug 07 2009

Interleaving of A007283 and A000079 without initial terms 1 and 2.
Equals A029744 without first two terms. Agrees with A145751 for all terms listed there (up to 65536).
Binomial transform is A078057 without initial 1, second binomial transform is A048580, third binomial transform is A163606, fourth binomial transform is A163604, fifth binomial transform is A163605.
a(n) is the number of vertices of the (n-1)-iterated line digraph L^{n-1}(G) of the digraph G(=L^0(G)) with vertices u,v,w and arcs u->v, v->u, v->w, w->v. - Miquel A. Fiol, Jun 08 2024

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Miquel A. Fiol, J. L. A. Yebra, and I. Alegre, Line digraph iterations and the (d,k) digraph problem, IEEE Trans. Comput. C-33(5) (1984), 400-403.
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A000079 (powers of 2), A007283 (3*2^n), A027383, A029744, A048580, A052955, A063759, A078057, A090989, A145751, A163604, A163605, A163606.

[ n le 2 select n+2 else 2*Self(n-2): n in [1..41] ];

LinearRecurrence[{0,2}, {3,4}, 52] (* or *) Table[(1/2)*(5-(-1)^n )*2^((2*n-1+(-1)^n)/4), {n,50}] (* G. C. Greubel, Aug 24 2017 *)

my(x='x+O('x^50)); Vec(x*(3+4*x)/(1-2*x^2)) \\ G. C. Greubel, Aug 24 2017

[(2+(n%2))*2^((n-(n%2))//2) for n in range(1,41)] # G. C. Greubel, Jun 13 2024

a(n) = A027383(n-1) + 2.
a(n) = A052955(n) + 1 for n >= 1.
a(n) = (1/2)*(5 - (-1)^n)*2^((2*n - 1 + (-1)^n)/4).
G.f.: x*(3+4*x)/(1-2*x^2).
a(n) = A090989(n-1).
E.g.f.: (1/2)*(4*cosh(sqrt(2)*x) + 3*sqrt(2)*sinh(sqrt(2)*x) - 4). - G. C. Greubel, Aug 24 2017
a(n) = A063759(n), n >= 1. - R. J. Mathar, Jan 25 2023

A208245 Triangle read by rows: a(n,k) = a(n-2,k) + a(n-2,k-1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 1, 1, 1, 3, 4, 3, 2, 1, 1, 1, 4, 6, 5, 3, 2, 1, 1, 1, 4, 7, 7, 5, 3, 2, 1, 1, 1, 5, 10, 11, 8, 5, 3, 2, 1, 1, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 1, 6, 15, 21, 19, 13, 8, 5, 3, 2, 1, 1, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1

Offset: 1

Richard R. Forberg, Apr 22 2013

Sum of terms in each row are given by sequence A052955.
Columns (at constant k) converge toward Fibonacci starting first from high value of k).
First seven rows are same as A008242. The odd numbered rows of this sequence equal the rows of A123736.  Also it has some similarities to A162741.
Columns (constant k), prior to convergence to Fibonacci, appear as various other sequences (e.g. k = 4, is sequence A055803, with other columns in same referenced family).

			The first 13 rows are (as above) where n is the row index:
1
1, 1
1, 1, 1
1, 2, 1, 1
1, 2, 2, 1, 1
1, 3, 3, 2, 1, 1
1, 3, 4, 3, 2, 1, 1
1, 4, 6, 5, 3, 2, 1, 1
1, 4, 7, 7, 5, 3, 2, 1, 1
1, 5, 10, 11, 8, 5, 3, 2, 1, 1
1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1
1, 6, 15, 21, 19, 13, 8, 5, 3, 2, 1, 1
1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1,

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened

Cf. A000045 (central terms).

a208245 n k = a208245_tabl !! (n-1) !! (k-1)
a208245_row n = a208245_tabl !! (n-1)
a208245_tabl = map fst $ iterate f ([1], [1, 1]) where
   f (us, vs) = (vs, zipWith (+) ([0] ++ us ++ [0]) (us ++ [0, 1]))
-- Reinhard Zumkeller, Jul 28 2013

a(n,k) = a(n-2,k) + a(n-2,k-1); if n=k or k=1 then a(n,k)=1; if n

A191795 Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k DUU's, where U=(1,1) and D=(1,-1).

Original entry on oeis.org

1, 1, 2, 3, 5, 1, 7, 3, 11, 9, 15, 19, 1, 23, 42, 5, 31, 77, 18, 47, 150, 54, 1, 63, 255, 137, 7, 95, 464, 333, 32, 127, 753, 720, 115, 1, 191, 1314, 1558, 360, 9, 255, 2067, 3067, 996, 50, 383, 3508, 6167, 2597, 214, 1, 511, 5397, 11410, 6207, 774, 11, 767, 8982, 21820, 14485, 2494, 72

Offset: 0

Emeric Deutsch, Jun 18 2011

Row n >= 1 contains ceiling(n/3) entries.
Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
T(n,0) = A052955(n-1).
Sum_{k>=0} k*T(n,k) = A191796(n).

			T(5,1)=3 because we have U(DUU)D, U(DUU)U, and UU(DUU), where U=(1,1) and D=(1,-1) (the DUU's are shown between parentheses).
Triangle starts:
   1;
   1;
   2;
   3;
   5,  1;
   7,  3;
  11,  9;
  15, 19, 1;
  23, 42, 5;

Cf. A001405, A052955, A191796.

eq := t*z^2*C^2-(1-2*z^2+2*t*z^2)*C+1-z^2+t*z^2 = 0: C := RootOf(eq, C): G := 1-(1-C-z*C)/(1-z+t*z-t*z*C): Gser := simplify(series(G, z = 0, 23)): for n from 0 to 18 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 18 do seq(coeff(P[n], t, k), k = 0 .. ceil((1/3)*n)-1) end do; # yields sequence in triangular form

G.f.: G(t,z) = 1 - (1-C-z*C)/(1-z+t*z-t*z*C), where C=C(t,z) is given by t*z^2*C^2 - (1-2*z^2+2*t*z^2)*C + 1-z^2+t*z^2 = 0.

A249452 Numbers k such that A249441(k) = 3.

Original entry on oeis.org

15, 31, 47, 63, 95, 127, 191, 255, 383, 511, 767, 1023, 1535, 2047, 3071, 4095, 6143, 8191, 12287, 16383, 24575, 32767, 49151, 65535, 98303, 131071, 196607, 262143, 393215, 524287, 786431, 1048575, 1572863, 2097151, 3145727, 4194303, 6291455, 8388607, 12582911

Offset: 1

Vladimir Shevelev, Oct 29 2014

Or k for which none of entries in the k-th row of Pascal's triangle (A007318) is divisible by 4 (cf. comment in A249441).
Using the Kummer carries theorem, one can prove that, for n>=2, a(n) has the form of either 1...1 or 101...1 in base 2.
The sequence is a subset of so-called binomial coefficient predictors (BCP) in base 2 (see Shevelev link, Th. 6 and Cor. 8), which were found also using Kummer theorem and have a very close binary structure.

E. E. Kummer, Über die Ergänzungssätze zu den allgemeinen Reciprocitätsgesetzen, J. Reine Angew. Math. 44 (1852), 93-146.
V. Shevelev, Binomial Coefficient Predictors, Journal of Integer Sequences, Vol. 14 (2011), Article 11.2.8
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Cf. A000225, A029744, A048278, A052955, A055010, A089633, A249441.

CoefficientList[Series[(15 + 16 x - 14 x^2 - 16 x^3)/(1 - x -2 x^2 + 2 x^3), {x, 0, 70}], x] (* Vincenzo Librandi, Oct 30 2014 *)
LinearRecurrence[{1,2,-2},{15,31,47,63},40] (* Harvey P. Dale, Apr 01 2019 *)

a(n)=if(n==1, 15, (n%2+2)<<(n\2+3)-1) \\ Charles R Greathouse IV, Nov 06 2014

is(n)=(n+1)>>valuation(n+1, 2)<5 && !setsearch([1, 2, 3, 5, 7, 11, 23], n) \\ Charles R Greathouse IV, Nov 06 2014

a(n) has either form 2^k - 1 or 3*2^m-1, k, m >= 4 (cf. A000225, A055010). Since, for k>=5, 2^k-1<3*2^(k-1)-1<2^(k+1)-1, we have that, for n>=1, a(2*n) = 2^(n+4)-1; a(2*n+1) = 3*2^(n+3)-1. - Vladimir Shevelev, Oct 29 2014, Nov 06 2014
a(1) = 15, and for n>1, a(n) = A052955(n+6). [Follows from above] - Antti Karttunen, Nov 03 2014
G.f.: (15+16*x-14*x^2-16*x^3)/(1-x-2*x^2+2*x^3); a(n) = 16*A029744(n)-1. - Peter J. C. Moses, Oct 30 2014

More terms from Peter J. C. Moses, Oct 29 2014

A364144 Number of distinct representations for n in base 2, using digits -1,0,1, whose sum of digits is 0.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 1, 3, 2, 4, 2, 4, 3, 4, 1, 3, 3, 5, 2, 6, 4, 6, 2, 5, 4, 7, 3, 6, 4, 5, 1, 3, 3, 6, 3, 7, 5, 8, 2, 7, 6, 10, 4, 10, 6, 8, 2, 6, 5, 9, 4, 10, 7, 10, 3, 8, 6, 10, 4, 8, 5, 6, 1, 3, 3, 6, 3, 8, 6, 9, 3, 8, 7, 13, 5, 12, 8, 11, 2, 8, 7, 13, 6

Offset: 0

Jeffrey Shallit, Jul 10 2023

			a(12) = 2, because 12 = 16-4 = 32-16-8+4.

Cf. A028310, A052955.

a364144(upto) = {my (a=vector(upto)); for (k=1, 3^floor(3*log(upto)), my (w=digits(k,3), n); w=apply(x->x-1, w); if (w[1] && vecsum(w)==0, my (n=fromdigits(w,b=2)); if (n>0 && n<=#a, a[n]++))); concat(1,a)};
a364144(70) \\ Hugo Pfoertner, Jul 11 2023

a(2^n) = 1.

a(2^n-1) = A028310(n).

a(0)=1 prepended by Alois P. Heinz, Jul 10 2023

A385396 Numbers k such that 8 does not divide binomial(k, j) for any j in 0..k.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 19, 23, 27, 31, 39, 47, 55, 63, 79, 95, 111, 127, 159, 191, 223, 255, 319, 383, 447, 511, 639, 767, 895, 1023, 1279, 1535, 1791, 2047, 2559, 3071, 3583, 4095, 5119, 6143, 7167, 8191, 10239, 12287, 14335, 16383, 20479, 24575

Offset: 1

Peter Luschny, Jun 28 2025

nonn

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,2,-2).

Cf. A000225 (case m=2), A052955 (case m=4).

isa := n -> andmap(j -> modp(binomial(n, j), 8) > 0, [seq(0..n)]): select(isa, [seq(0..200)]);
# Or, using the o.g.f.:
gf := (x + x^2 + x^3 + x^4 - x^5 - x^6 - x^7)/((-1 + x)*(-1 + 2*x^4)): ser := series(gf, x, 60): seq(coeff(ser, x, n), n = 0..53);

LinearRecurrence[{1, 0, 0, 2, -2}, Range[0, 7], 60] (* Paolo Xausa, Jun 30 2025 *)

def seq_gen():
    n, c, value = 0, 1, 3
    for v in [0, 1, 2]: yield v
    while True:
        yield value
        value += c
        n += 1
        if n == 4:
            n = 0
            c += c
term = seq_gen()
print([next(term) for _ in range(54)])

a(n) = [x^n] (x + x^2 + x^3 + x^4 - x^5 - x^6 - x^7)/((-1 + x)*(-1 + 2*x^4)).

a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) for n > 8. - Chai Wah Wu, Jun 28 2025

A065548 a(1) = 1 and, for n > 0, a(2n) = sum{a(i) | 0 < i <= n}, a(2n+1) = a(n)^2.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 5, 4, 6, 1, 9, 9, 10, 1, 15, 25, 19, 16, 25, 36, 26, 1, 35, 81, 44, 81, 54, 100, 55, 1, 70, 225, 95, 625, 114, 361, 130, 256, 155, 625, 191, 1296, 217, 676, 218, 1, 253, 1225, 334, 6561, 378, 1936, 459, 6561, 513, 2916, 613, 10000, 668, 3025, 669, 1

Offset: 1

Reinhard Zumkeller, Nov 28 2001

nonn

a(A052955(k)) = 1 for all k.

A216344 Triangle T(n,k), read by rows, given by (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, -1, 1, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938 .

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 4, 3, 1, 0, 8, 8, 7, 4, 1, 0, 16, 16, 16, 11, 5, 1, 0, 32, 32, 36, 28, 16, 6, 1, 0, 64, 64, 80, 68, 45, 22, 7, 1, 0, 128, 128, 176, 160, 118, 68, 29, 8, 1, 0, 256, 256, 384

Offset: 0

Philippe Deléham, Sep 04 2012

			Triangle begins :
1
0, 1
0, 1, 1
0, 2, 2, 1
0, 4, 4, 3, 1
0, 8, 8, 7, 4, 1
0, 16, 16, 16, 11, 5, 1
0, 32, 32, 36, 28, 16, 6, 1

Cf. A034943

G.f.: (1-2*x+y*x^2)/(1-2*x-y*x+2*y*x^2-y^2*x^3)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k-1) + T(n-3,k-2), T(0,0) = T(1,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0 and T(n,k) = 0 if k<0 or if k>n .
Sum_{k, 0<=k<=n} T(n,k) = A034943(n+1) .
Sum_{k, 0<=k<=n} T(n,k)*2^k*(-1/2)^(n-k) = A052955(n) .
T(n+1,1) = A011782(n), T(n+2,2) = 2^n = A000079(n), T(n+3,3) = A045891(n+1) .

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A163978 a(n) = 2*a(n-2) for n > 2; a(1) = 3, a(2) = 4.

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A332954 Triangle read by rows: T(n,k) is the number of permutations sigma of [n] such that sigma(j)/(j+k) > sigma(j+1)/(j+k+1) for 1 <= j <= n-1.

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A206918 Sum of binary palindromes p < 2^n.

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A208245 Triangle read by rows: a(n,k) = a(n-2,k) + a(n-2,k-1).

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A191795 Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k DUU's, where U=(1,1) and D=(1,-1).

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A249452 Numbers k such that A249441(k) = 3.

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A364144 Number of distinct representations for n in base 2, using digits -1,0,1, whose sum of digits is 0.

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A385396 Numbers k such that 8 does not divide binomial(k, j) for any j in 0..k.

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A065548 a(1) = 1 and, for n > 0, a(2n) = sum{a(i) | 0 < i <= n}, a(2n+1) = a(n)^2.