This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(4) = 23 = 1*1 + 2*2 + 3*2 + 4*12 = (1 + 4 + 6 + 12), given Golomb's sequence (1, 2, 2, 3, 3, 4, 4, 4,...). a(4) = 23 = sum of row 4 terms of triangle A143124: (8 + 7 + 5 + 3).
Since Golomb's sequence starts 1,2,2,3,3,4,4,4,5,... the first run of length three starts at position 6, so a(3) = 6.
a[1] = 1; a[n_] := a[n] = 1 + a[n - a[a[n - 1]]]; s = Array[a, {13000}]; t = Most@ Map[Last, Tally@ s]; Table[Total@ Take[t, First@ Flatten@ Position[t, n]] - n + 1, {n, 50}] (* Michael De Vlieger, Oct 07 2015, after Robert G. Wilson v at A001462 *)
a(n) = if (n<2, 1, a(n-1)+ t(n-1)*(n-1)); t(n) = local(A, t, i); if(n<3, max(0, n), A=vector(n); t=A[i=2]=2; for(k=3, n, A[k]=A[k-1]+if(t--==0, t=A[i++ ]; 1)); A[n]); vector(100, n, a(n))\\ Altug Alkan, Oct 06 2015
Golomb's sequence begins 1, 2,2, 3,3, 4,4,4, 5,5,5, ... and we just extend each run by one term, getting 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, ...
The first terms, alongside the corresponding phrases, are: n a(n) Corresponding phrases -- ---- --------------------- 1 1 1 2 1 2 3 2 2, 3 4 1 3 5 1 4 6 2 4, 4 7 1 5 8 2 5, 5 9 1 6 10 2 6, 6 11 2 6, 7 12 1 7 13 2 7, 7 14 1 8 15 2 8, 8
\\ See Links section.
1.4107845307495355919347994202105751786146865173661...
a[1] = 1; a[n_] := a[n] = 1 + a[n - a[a[n - 1]]]; (* A001462 *) GenA385303[n_Integer] := Module[{cf1, cf2, d1, d2, i = n}, While[i < 2 n, cf1 = Table[a[k], {k, 1, i}]; cf2 = Table[a[k], {k, 1, i + 1}]; d1 = RealDigits[FromContinuedFraction[cf1], 10, n+1][[1]]; d2 = RealDigits[FromContinuedFraction[cf2], 10, n+1][[1]]; If[Take[d1,n] === Take[d2,n], Return[Take[d1,n]]]; i++;]]; GenA385303[100]
G.f. = x + x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + ... As triangle, the sequence starts 0; 1, 1; 2, 2, 2; 3, 3, 3, 3; 4, 4, 4, 4, 4; 5, 5, 5, 5, 5, 5; 6, 6, 6, 6, 6, 6, 6; 7, 7, 7, 7, 7, 7, 7, 7; 8, 8, 8, 8, 8, 8, 8, 8, 8; ...
a003056 = floor . (/ 2) . (subtract 1) .
sqrt . (+ 1) . (* 8) . fromIntegral
a003056_row n = replicate (n + 1) n
a003056_tabl = map a003056_row [0..]
a003056_list = concat $ a003056_tabl
-- Reinhard Zumkeller, Aug 02 2014, Oct 17 2010
[Floor((Sqrt(1+8*n)-1)/2): n in [0..80]]; // Vincenzo Librandi, Oct 23 2011
A003056 := (n,k) -> n: # Peter Luschny, Oct 29 2011 a := [ 0 ]: for i from 1 to 15 do for j from 1 to i+1 do a := [ op(a),i ]; od: od: a; A003056 := proc(n) floor((sqrt(1+8*n)-1)/2) ; end proc: # R. J. Mathar, Jul 10 2015
f[n_] := Floor[(Sqrt[1 + 8n] - 1)/2]; Table[ f[n], {n, 0, 87}] (* Robert G. Wilson v, Oct 21 2005 *)
Table[x, {x, 0, 13}, {y, 0, x}] // Flatten
T[ n_, k_] := If[ n >= k >= 0, n, 0]; (* Michael Somos, Dec 22 2016 *)
Flatten[Table[PadRight[{},n+1,n],{n,0,12}]] (* Harvey P. Dale, Jul 03 2021 *)
A003056(n)=(sqrtint(8*n+1)-1)\2 \\ M. F. Hasler, Oct 08 2011
t1(n)=floor(-1/2+sqrt(2+2*n)) /* A003056 */
t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */
from math import isqrt def A003056(n): return (k:=isqrt(m:=n+1<<1))+int((m<<2)>(k<<2)*(k+1)+1)-1 # Chai Wah Wu, Jul 26 2022
Start with a(1) = 1. By definition of the sequence, this says that the first run has length 1, so it must be a single 1, and a(2) = 2. Thus, the second run (which starts with this 2) must have length 2, so the third term must be also be a(3) = 2, and the fourth term can't be a 2, so must be a(4) = 1. Since a(3) = 2, the third run must have length 2, so we deduce a(5) = 1, a(6) = 2, and so on. The correction I made was to change a(4) to a(5) and a(5) to a(6). - _Labos Elemer_, corrected by _Graeme McRae_
a = 1:2: drop 2 (concat . zipWith replicate a . cycle $ [1,2]) -- John Tromp, Apr 09 2011
M := 100; s := [ 1,2,2 ]; for n from 3 to M do for i from 1 to s[ n ] do s := [ op(s),1+((n-1)mod 2) ]; od: od: s; A000002 := n->s[n]; # alternative implementation based on the Cloitre formula: A000002 := proc(n) local ksu,k ; option remember; if n = 1 then 1; elif n <=3 then 2; else for k from 1 do ksu := add(procname(i),i=1..k) ; if n = ksu then return (3+(-1)^k)/2 ; elif n = ksu+ 1 then return (3-(-1)^k)/2 ; end if; end do: end if; end proc: # R. J. Mathar, Nov 15 2014
a[steps_] := Module[{a = {1, 2, 2}}, Do[a = Append[a, 1 + Mod[(n - 1), 2]], {n, 3, steps}, {i, a[[n]]}]; a]
a[ n_] := If[ n < 3, Max[ 0, n], Module[ {an = {1, 2, 2}, m = 3}, While[ Length[ an] < n, an = Join[ an, Table[ Mod[m, 2, 1], { an[[ m]]} ]]; m++]; an[[n]]]] (* Michael Somos, Jul 11 2011 *)
n=8; Prepend[ Nest[ Flatten[ Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, n], 1] (* Birkas Gyorgy, Jul 10 2012 *)
KolakoskiSeq[n_Integer] := Block[{a = {1, 2, 2}}, Fold[Join[#1, ConstantArray[Mod[#2, 2, 1], #1[[#2]]]] &, a, Range[3, n]]]; KolakoskiSeq[999] (* Mikk Heidemaa, Nov 01 2024 *) (* Corrected by Giorgos Kalogeropoulos, May 09 2025 *)
my(a=[1,2,2]); for(n=3,80, for(i=1,a[n],a=concat(a,2-n%2))); a
{a(n) = local(an=[1, 2, 2], m=3); if( n<1, 0, while( #an < n, an = concat( an, vector(an[m], i, 2-m%2)); m++); an[n])};
# For explanation see link.
def Kolakoski():
x = y = -1
while True:
yield [2,1][x&1]
f = y &~ (y+1)
x ^= f
y = (y+1) | (f & (x>>1))
K = Kolakoski()
print([next(K) for in range(100)]) # _David Eppstein, Oct 15 2016
From _Clark Kimberling_, Sep 16 2008: (Start) As a rectangular array, a northwest corner: 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 This is the weight array (cf. A144112) of A107985 (formatted as a rectangular array). (End) G.f. = x + 2*x^2 + 2*x^3 + 3*x^4 + 3*x^5 + 3*x^6 + 4*x^7 + 4*x^9 + 4*x^9 + 4*x^10 + ...
a002024 n k = a002024_tabl !! (n-1) !! (k-1) a002024_row n = a002024_tabl !! (n-1) a002024_tabl = iterate (\xs@(x:_) -> map (+ 1) (x : xs)) [1] a002024_list = concat a002024_tabl a002024' = round . sqrt . (* 2) . fromIntegral -- Reinhard Zumkeller, Jul 05 2015, Feb 12 2012, Mar 18 2011
a002024_list = [1..] >>= \n -> replicate n n
a002024 = (!!) $ [1..] >>= \n -> replicate n n -- Sascha Mücke, May 10 2016
[Floor(Sqrt(2*n) + 1/2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2014
A002024 := n-> ceil((sqrt(1+8*n)-1)/2); seq(A002024(n), n=1..100);
a[1] = 1; a[n_] := a[n] = a[n - a[n - 1]] + 1 (* Branko Curgus, May 12 2009 *) Table[n, {n, 13}, {n}] // Flatten (* Robert G. Wilson v, May 11 2010 *) Table[PadRight[{},n,n],{n,15}]//Flatten (* Harvey P. Dale, Jan 13 2019 *)
t1(n)=floor(1/2+sqrt(2*n)) /* A002024 = this sequence */
t2(n)=n-binomial(floor(1/2+sqrt(2*n)),2) /* A002260(n-1) */
t3(n)=binomial(floor(3/2+sqrt(2*n)),2)-n+1 /* A004736 */
t4(n)=n-1-binomial(floor(1/2+sqrt(2*n)),2) /* A002260(n-1)-1 */
A002024(n)=(sqrtint(n*8)+1)\2 \\ M. F. Hasler, Apr 19 2014
a(n)=(sqrtint(8*n-7)+1)\2
a(n)=my(k=1);while(binomial(k+1,2)+1<=n,k++);k \\ R. J. Cano, Mar 17 2014
from math import isqrt def A002024(n): return (isqrt(8*n)+1)//2 # Chai Wah Wu, Feb 02 2022
[floor(sqrt(2*n) +1/2) for n in (1..80)] # G. C. Greubel, Dec 10 2018
p_{4*square}(23)=1 because 23 = 3^2 + 3^2 + 2^2 + 1^2 and there is no other partition of 23 into squares.
G.f.: A(x) = 1 + x + x^2 + x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 +...
such that the g.f. A(x) satisfies the identity [_Paul D. Hanna_]:
A(x) = 1/((1-x)*(1-x^4)*(1-x^9)*(1-x^16)*(1-x^25)*...)
A(x) = 1 + x/(1-x) + x^4/((1-x)*(1-x^4)) + x^9/((1-x)*(1-x^4)*(1-x^9)) + x^16/((1-x)*(1-x^4)*(1-x^9)*(1-x^16)) + ...
From _Gus Wiseman_, Mar 09 2019: (Start)
The a(14) = 6 integer partitions into squares are:
(941)
(911111)
(44411)
(44111111)
(41111111111)
(11111111111111)
while the a(14) = 6 integer partitions in which the multiplicity of k is a multiple of k for all k are:
(333221)
(33311111)
(22222211)
(2222111111)
(221111111111)
(11111111111111)
(End)
a001156 = p (tail a000290_list) where p _ 0 = 1 p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m -- Reinhard Zumkeller, Oct 31 2012, Aug 14 2011
m:=70; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (&*[1/(1-x^(k^2)): k in [1..(m+2)]]) )); // G. C. Greubel, Nov 11 2018
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
b(n, i-1)+ `if`(i^2>n, 0, b(n-i^2, i))))
end:
a:= n-> b(n, isqrt(n)):
seq(a(n), n=0..120); # Alois P. Heinz, May 30 2014
CoefficientList[ Series[Product[1/(1 - x^(m^2)), {m, 70}], {x, 0, 68}], x] (* Or *)
Join[{1}, Table[Length@PowersRepresentations[n, n, 2], {n, 68}]] (* Robert G. Wilson v, Apr 12 2005, revised Sep 27 2011 *)
f[n_] := Length@ IntegerPartitions[n, All, Range@ Sqrt@ n^2]; Array[f, 67] (* Robert G. Wilson v, Apr 14 2013 *)
b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, b[n, i-1] + If[i^2>n, 0, b[n-i^2, i]]]]; a[n_] := b[n, Sqrt[n]//Floor]; Table[a[n], {n, 0, 120}] (* Jean-François Alcover, Nov 02 2015, after Alois P. Heinz *)
{a(n)=polcoeff(1/prod(k=1, sqrtint(n+1), 1-x^(k^2)+x*O(x^n)), n)} \\ Paul D. Hanna, Mar 09 2012
{a(n)=polcoeff(1+sum(m=1, sqrtint(n+1), x^(m^2)/prod(k=1, m, 1-x^(k^2)+x*O(x^n))), n)} \\ Paul D. Hanna, Mar 09 2012
a(50)=3 because we have [1,4,9,36], [1,49], and [9,16,25]. - _Emeric Deutsch_, Jan 26 2016 From _Gus Wiseman_, Mar 09 2019: (Start) The first 30 terms count the following integer partitions: 1: (1) 4: (4) 5: (4,1) 9: (9) 10: (9,1) 13: (9,4) 14: (9,4,1) 16: (16) 17: (16,1) 20: (16,4) 21: (16,4,1) 25: (25) 25: (16,9) 26: (25,1) 26: (16,9,1) 29: (25,4) 29: (16,9,4) 30: (25,4,1) 30: (16,9,4,1) (End)
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
b(n, i-1) +`if`(i^2>n, 0, b(n-i^2, i-1))))
end:
a:= n-> b(n, isqrt(n)):
seq(a(n), n=0..100); # Alois P. Heinz, May 14 2014
nn=10; CoefficientList[Series[Product[(1+x^(k*k)), {k,nn}], {x,0,nn*nn}], x] (* T. D. Noe, Jul 24 2006 *)
b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, b[n, i-1] + If[i^2 > n, 0, b[n - i^2, i-1]]]]; a[n_] := b[n, Floor[Sqrt[n]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Sep 21 2015, after Alois P. Heinz *)
nmax = 20; poly = ConstantArray[0, nmax^2 + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k^2 + 1]], {j, nmax^2, k^2, -1}];, {k, 2, nmax}]; poly (* Vaclav Kotesovec, Dec 09 2016 *)
Table[Length[Select[IntegerPartitions[n],Reverse[Union[#]]==Length/@Split[#]&]],{n,30}] (* Gus Wiseman, Mar 09 2019 *)
a(n)=polcoeff(prod(k=1,sqrt(n),1+x^k^2), n)
first(n)=Vec(prod(k=1,sqrtint(n),1+'x^k^2,O('x^(n+1))+1)) \\ Charles R Greathouse IV, Sep 03 2015
from functools import cache from sympy.core.power import isqrt @cache def b(n,i): # Code after Alois P. Heinz if n == 0: return 1 if i == 0: return 0 i2 = i*i return b(n, i-1) + (0 if i2 > n else b(n - i2, i-1)) a = lambda n: b(n, isqrt(n)) print([a(n) for n in range(1, 101)]) # Darío Clavijo, Nov 30 2023
Comments