cp's OEIS Frontend

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.

User: Christopher Carl Heckman

Christopher Carl Heckman's wiki page.

Christopher Carl Heckman has authored 4 sequences.

A225378 Construct sequences P,Q,R by the rules: Q = first differences of P, R = second differences of P, P starts with 1,5,11, Q starts with 4,6, R starts with 2; at each stage the smallest number not yet present in P,Q,R is appended to R; every number appears exactly once in the union of P,Q,R. Sequence gives R.

Original entry on oeis.org

2, 3, 7, 8, 10, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 61, 62, 63, 64
Offset: 1

Views

Author

N. J. A. Sloane, May 12 2013, based on email from Christopher Carl Heckman, May 06 2013

Keywords

Comments

P can be extended for 10^6 terms, but it is not known if P,Q,R can be extended to infinity.
A probabilistic argument suggests that P, Q, R are infinite. - N. J. A. Sloane, May 19 2013

Examples

			The initial terms of P, Q, R are:
1     5    11    20    36    60    94   140   199   272   360
   4     6     9    16    24    34    46    59    73    88
      2     3     7     8    10    12    13    14    15

Links

Crossrefs

Cf. A225376, A225377, A005228, A030124, A037257.

Programs

Extensions

Corrected and edited by Christopher Carl Heckman, May 12 2013

A225377 Construct sequences P,Q,R by the rules: Q = first differences of P, R = second differences of P, P starts with 1,5,11, Q starts with 4,6, R starts with 2; at each stage the smallest number not yet present in P,Q,R is appended to R; every number appears exactly once in the union of P,Q,R. Sequence gives Q.

Original entry on oeis.org

4, 6, 9, 16, 24, 34, 46, 59, 73, 88, 105, 123, 142, 163, 185, 208, 233, 259, 286, 314, 343, 373, 404, 436, 469, 504, 541, 579, 618, 658, 699, 741, 784, 828, 873, 920, 968, 1017, 1067, 1118, 1170
Offset: 1

Views

Author

N. J. A. Sloane, May 12 2013, based on email from Christopher Carl Heckman, May 06 2013

Keywords

Comments

P can be extended for 10^6 terms, but it is not known if P,Q,R can be extended to infinity.
A probabilistic argument suggests that P, Q, R are infinite. - N. J. A. Sloane, May 19 2013

Examples

			The initial terms of P, Q, R are:
1     5    11    20    36    60    94   140   199   272   360
   4     6     9    16    24    34    46    59    73    88
      2     3     7     8    10    12    13    14    15

Links

Crossrefs

Cf. A225376, A225378, A005228, A030124, A037257.

Programs

Extensions

Corrected and edited by Christopher Carl Heckman, May 12 2013

A225376 Construct sequences P,Q,R by the rules: Q = first differences of P, R = second differences of P, P starts with 1,5,11, Q starts with 4,6, R starts with 2; at each stage the smallest number not yet present in P,Q,R is appended to R; every number appears exactly once in the union of P,Q,R. Sequence gives P.

Original entry on oeis.org

1, 5, 11, 20, 36, 60, 94, 140, 199, 272, 360, 465, 588, 730, 893, 1078, 1286, 1519, 1778, 2064, 2378, 2721, 3094, 3498, 3934, 4403, 4907, 5448, 6027, 6645, 7303, 8002, 8743, 9527, 10355, 11228
Offset: 1

Views

Author

N. J. A. Sloane, May 12 2013, based on email from Christopher Carl Heckman, May 06 2013

Keywords

Comments

P can be extended for 10^6 terms, but it is not known if P,Q,R can be extended to infinity.
A probabilistic argument suggests that P, Q, R are infinite. - N. J. A. Sloane, May 19 2013
Martin Gardner (see reference) states that no such triple P,Q,R of sequences exists if it is required that P(1)

Examples

			The initial terms of P, Q, R are:
1     5    11    20    36    60    94   140   199   272   360
   4     6     9    16    24    34    46    59    73    88
      2     3     7     8    10    12    13    14    15

References

  • M. Gardner, Weird Numbers from Titan, Isaac Asimov's Science Fiction Magazine, Vol. 4, No. 5, May 1980, pp. 42ff.

Links

Crossrefs

Cf. A225377, A225378, A005228, A030124, A037257.

Programs

  • Maple
    Hofstadter2 := proc (N) local h, dh, ddh, S, lbmex, i:
    h := 1, 5, 11: dh := 4, 6: ddh := 2:
    lbmex := 3: S := {h,dh,ddh}:
    for i from 4 to N do:
       while lbmex in S do: S := S minus {lbmex}: lbmex := lbmex + 1: od:
       ddh := ddh, lbmex:
       dh := dh, dh[-1] + lbmex:
       h := h, h[-1] + dh[-1]:
       S := S union {h[-1], dh[-1], ddh[-1]}:
       lbmex := lbmex + 1:
    od:
    if {h} intersect {dh} <> {} then: return NULL:
    elif {h} intersect {ddh} <> {} then: return NULL:
    elif {ddh} intersect {dh} <> {} then: return NULL:
    else: return [h]: fi:
end proc: # Christopher Carl Heckman, May 12 2013
  • Mathematica
    Hofstadter2[N_] := Module[{P, Q, R, S, k, i}, P = {1, 5, 11}; Q = {4, 6}; R = {2}; k = 3; S = Join[P, Q, R]; For[i = 4, i <= N, i++, While[MemberQ[S, k], S = S~Complement~{k}; k++]; AppendTo[R, k]; AppendTo[Q, Q[[-1]] + k]; AppendTo[P, P[[-1]] + Q[[-1]]]; S = S~Union~{P[[-1]], Q[[-1]], R[[-1]]}; k++]; Which[P~Intersection~Q != {}, Return@Nothing, {P}~Intersection~R != {}, Return@Nothing, R~Intersection~Q != {}, Return@Nothing, True, Return@P]];
Hofstadter2[36] (* Jean-François Alcover, Mar 05 2023, after Christopher Carl Heckman's Maple code *)

Extensions

Corrected and edited by Christopher Carl Heckman, May 12 2013

A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.

Original entry on oeis.org

1, 2, 2, 3, 6, 9, 15, 30, 54, 97, 189, 360, 675, 1304, 2522, 4835, 9358, 18193, 35269, 68568, 133737, 260802, 509132, 995801, 1948931, 3816904, 7483636, 14683721, 28827798, 56637969, 111347879, 219019294, 431043814, 848764585, 1672056525, 3295390800, 6497536449
Offset: 0

Views

Author

Christopher Carl Heckman, Jul 29 2009

Keywords

Comments

A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
Proposed Problem 11610 in the Dec 2011 A.M.M.
From Gus Wiseman, Jul 27 2021: (Start)
Also the antidiagonal sums of the matrices counting integer compositions by length and alternating sum (A345197). So a(n) is the number of integer compositions of n + 1 of length (n - s + 3)/2, where s is the alternating sum of the composition. For example, the a(0) = 1 through a(6) = 7 compositions are:
(1) (2) (3) (4) (5) (6) (7)
(11) (21) (31) (41) (51) (61)
(121) (122) (123) (124)
(221) (222) (223)
(1112) (321) (322)
(1211) (1122) (421)
(1221) (1132)
(2112) (1231)
(2211) (2122)
(2221)
(3112)
(3211)
(11131)
(12121)
(13111)
For a bijection with the main (binary string) interpretation, take the run-lengths of each binary string of length n + 1 that satisfies the condition and starts with 1.
(End)

Examples

			1 + 2*x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 15*x^6 + 30*x^7 + 54*x^8 + 97*x^9 + ...
From _Gus Wiseman_, Jul 27 2021: (Start)
The a(0) = 1 though a(6) = 15 binary strings:
  ()  (0)  (1,0)  (0,0,1)  (0,0,1,0)  (0,0,1,1,0)  (0,0,0,1,0,1)
      (1)  (1,1)  (1,1,0)  (0,0,1,1)  (0,0,1,1,1)  (0,0,1,0,0,1)
                  (1,1,1)  (0,1,0,0)  (0,1,1,0,0)  (0,0,1,1,1,0)
                           (1,0,0,1)  (1,0,0,1,0)  (0,0,1,1,1,1)
                           (1,1,1,0)  (1,0,0,1,1)  (0,1,0,0,0,1)
                           (1,1,1,1)  (1,0,1,0,0)  (0,1,1,1,0,0)
                                      (1,1,0,0,1)  (1,0,0,1,1,0)
                                      (1,1,1,1,0)  (1,0,0,1,1,1)
                                      (1,1,1,1,1)  (1,0,1,1,0,0)
                                                   (1,1,0,0,1,0)
                                                   (1,1,0,0,1,1)
                                                   (1,1,0,1,0,0)
                                                   (1,1,1,0,0,1)
                                                   (1,1,1,1,1,0)
                                                   (1,1,1,1,1,1)
(End)

Links

Crossrefs

Antidiagonal sums of the matrices A345197.
Row sums of A345907.
Taking diagonal instead of antidiagonal sums gives A345908.
A011782 counts compositions (or binary strings).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.
Cf. A000041, A000070, A000096, A000097, A000124, A000346, A007318, A008549, A025047, A131577, A238279.

Programs

  • Maple
    with(combinat): count := proc(n) local S, matches, A, k, i; S := subsets(\{seq(i, i=1..n)\}): matches := 0: while not S[finished] do A := S[nextvalue](): k := 0: for i from 1 to n-1 do: if not (i in A) and not (i+1 in A) then k := k + 1: fi: if not (i in A) and (i+1 in A) then k := k - 1: fi: od: if (k = 0) then matches := matches + 1: fi: end do; return(matches); end proc:
# second Maple program:
b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 1, 0):
seq(a(n), n=0..36);  # Alois P. Heinz, Mar 20 2024
  • Mathematica
    a[0] = 1; a[n_] := Sum[Binomial[2*k - 1, k]*Binomial[n - 2*k, k] + Binomial[2*k, k]*Binomial[n - 2*k - 1, k], {k, 0, n/3}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 28 2017, after Joel B. Lewis *)
Table[Length[Select[Tuples[{0,1},n],Count[Partition[#,2,1],{0,0}]==Count[Partition[#,2,1],{0,1}]&]],{n,0,10}] (* Gus Wiseman, Jul 27 2021 *)
a[0]:=1; a[n_]:=(1 + 3*HypergeometricPFQ[{1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3},{1, (1-n)/2, 1-n/2, -3*n/8}, -27])/2; Array[a,37,0] (* Stefano Spezia, Apr 26 2024 *)
  • Python
    from math import comb
def A163493(n): return 2+sum((x:=comb((k:=m<<1)-1,m)*comb(n-k,m))+(x*(n-3*m)<<1)//(n-k) for m in range(1,n//3+1)) if n else 1 # Chai Wah Wu, May 01 2024

Formula

G.f.: 1/2/(1-x) + (1+2*x)/2/sqrt((1-x)*(1-2*x)*(1+x+2*x^2)). - Richard Stanley, corrected Apr 29 2011
G.f.: (1 + sqrt( 1 + 4*x / ((1 - x) * (1 - 2*x) * (1 + x + 2*x^2)))) / (2*(1 - x)). - Michael Somos, Jan 30 2012
a(n) = sum( binomial(2*k-1, k)*binomial(n-2*k,k) + binomial(2*k, k)*binomial(n-2*k-1, k), k=0..floor(n/3)). - Joel B. Lewis, May 21 2011
Conjecture: -n*a(n) +(2+n)*a(n-1) +(3n-12)*a(n-2) +(12-n)*a(n-3) +(2n-18)*a(n-4)+(56-12n)*a(n-5) +(8n-40)*a(n-6)=0. - R. J. Mathar, Nov 28 2011
G.f. y = A(x) satisfies x = (1 - x) * (1 - 2*x) * (1 + x + 2*x^2) * y * (y * (1 - x) - 1). - Michael Somos, Jan 30 2012
Sequence a(n) satisfies 0 = a(n) * (n^2-2*n) + a(n-1) * (-3*n^2+8*n-2) + a(n-2) * (3*n^2-10*n+2) + a(n-3) * (-5*n^2+18*n-6) + a(n-4) * (8*n^2-34*n+22) + a(n-5) * (-4*n^2+20*n-16) except if n=1 or n=2. - Michael Somos, Jan 30 2012
a(n) = (1 + 3*hypergeom([1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3],[1, (1-n)/2, 1-n/2, -3*n/8],-27))/2 for n > 0. - Stefano Spezia, Apr 26 2024
a(n) ~ 2^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 26 2024

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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