A035106 -id:A035106 - cp's OEIS frontend

A163253 An interspersion: the order array of the odd-numbered columns of the double interspersion at A161179.

1, 4, 2, 9, 5, 3, 16, 10, 7, 6, 25, 17, 13, 11, 8, 36, 26, 21, 18, 14, 12, 49, 37, 31, 27, 22, 19, 15, 64, 50, 43, 38, 32, 28, 23, 20, 81, 65, 57, 51, 44, 39, 33, 29, 24, 100, 82, 73, 66, 58, 52, 45, 40, 34, 30, 121, 101, 91, 83, 74, 67, 59, 53, 46, 41, 35

Offset: 1

Clark Kimberling, Jul 23 2009

A permutation of the natural numbers.
Row 1 consists of the squares.
Beginning with row 5, the columns obey a 3rd-order recurrence:
c(n)=c(n-1)+c(n-2)-c(n-3)+1; thus disregarding row 1, the nonsquares are partitioned by this recurrence.
Except for initial terms, the first ten rows match A000290, A002522, A002061, A059100, A014209, A117950, A027688, A087475, A027689, A117951, and the first column, A035106.

			Corner:
1....4....9...16...25
2....5...10...17...26
3....7...13...21...31
6...11...18...27...38
The double interspersion A161179 begins thus:
1....4....7...12...17...24
2....3....8...11...18...23
5....6...13...16...25...30
9...10...19...22...33...38
Expel the even-numbered columns, leaving
1....7...17...
2....8...18...
5...13...25...
9...19...33...
Then replace each of those numbers by its rank when all the numbers are jointly ranked.

Clark Kimberling, Doubly interspersed sequences, double interspersions and fractal sequences, The Fibonacci Quarterly 48 (2010) 13-20.

Cf. A000037, A000290, A161179, A163254, A163255, A163256, A163257, A163258.

Let S(n,k) denote the k-th term in the n-th row. Three cases:
S(1,k)=k^2;
if n is even, then S(n,k)=k^2+(n-2)k+(n^2-2*n+4)/4;
if n>=3 is odd, then S(n,k)=k^2+(n-2)k+(n^2-2*n+1)/4.

Edited and augmented by Clark Kimberling, Jul 24 2009

A035104 First differences give (essentially) A028242.

Original entry on oeis.org

1, 4, 9, 13, 19, 24, 31, 37, 45, 52, 61, 69, 79, 88, 99, 109, 121, 132, 145, 157, 171, 184, 199, 213, 229, 244, 261, 277, 295, 312, 331, 349, 369, 388, 409, 429, 451, 472, 495, 517, 541, 564, 589, 613, 639, 664, 691, 717, 745, 772, 801, 829, 859, 888, 919

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A004652, A035106, A035107.

[(5+3*(-1)^n+28*n+2*n^2)/8: n in [0..60]]; // Vincenzo Librandi, Oct 20 2013

CoefficientList[Series[(3 x^3 - x^2 - 2 x - 1)/((x - 1)^3 (x + 1)), {x, 0, 60}], x] (* Vincenzo Librandi, Oct 20 2013 *)

From  Colin Barker, Mar 04 2013: (Start)
a(n) = (5+3*(-1)^n+28*n+2*n^2)/8.
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4).
G.f.: (3*x^3-x^2-2*x-1) / ((x-1)^3*(x+1)). (End)
Sum_{n>=0} 1/a(n) = 983/990 + tan(3*sqrt(5)*Pi/2)*Pi/(3*sqrt(5)) - cot(2*sqrt(3)*Pi)*Pi/(4*sqrt(3)). - Amiram Eldar, Sep 24 2022

More terms from Vincenzo Librandi, Oct 20 2013

A064797 Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies lcm(p_i, p_{i+1}) >= m for some i, 1 <= i <= n, where p_{n+1} = p_1.

Original entry on oeis.org

1, 2, 6, 6, 12, 12, 15, 15, 18, 18, 24, 24, 35, 35, 35, 35, 44, 44, 55, 55, 55, 55, 68, 68, 68, 68, 68, 68, 85, 85, 102, 102, 102, 102, 102, 102, 119, 119, 119, 119, 145, 145, 174, 174, 174, 174, 203, 203, 203, 203, 203, 203, 232, 232, 232, 232, 232, 232, 261, 261

Offset: 1

N. J. A. Sloane, Oct 21 2001

Testing a trial value of a(n) is equivalent to searching for a Hamiltonian cycle in the appropriate graph. - Martin Fuller, Jul 30 2006

			n=4: we must arrange the numbers 1..4 in a circle so that the max of the lcm of pairs of adjacent terms is minimized. The answer is 1423, with max lcm = 6, so a(4) = 6.

Cf. A064764, A035106, A064796, A000720, A073818.

Table[Min[Max[LCM@@@Partition[#,2,1,1]]&/@Permutations[Range[n]]], {n,10}] (* Harvey P. Dale, Oct 05 2011 *) (* The program takes a long time to run and uses a great deal of memory *)

For n >= 3, a(n) >= A073818(pi(n)+1), with equality for 17 <= n <= 250 - Martin Fuller, Jul 30 2006

More terms from Vladeta Jovovic, Oct 22 2001
a(11)-a(24) from Charles R Greathouse IV, Jul 23 2006
More terms from Martin Fuller, Jul 30 2006

A035107 First differences give (essentially) A028242.

Original entry on oeis.org

3, 9, 17, 29, 44, 64, 88, 118, 153, 195, 243, 299, 362, 434, 514, 604, 703, 813, 933, 1065, 1208, 1364, 1532, 1714, 1909, 2119, 2343, 2583, 2838, 3110, 3398, 3704, 4027, 4369, 4729, 5109, 5508, 5928, 6368, 6830, 7313, 7819, 8347, 8899, 9474

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A028242, A004652, A035104, A035106.

[(4*n^3+54*n^2+212*n+153-9*(-1)^n)/48: n in [0..50]]; // Vincenzo Librandi, Oct 21 2013

LinearRecurrence[{3,-2,-2,3,-1},{3,9,17,29,44},50] (* Harvey P. Dale, Oct 20 2013 *)
CoefficientList[Series[(2 x^3 - 4 x^2 + 3)/((x - 1)^4 (x + 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 21 2013 *)

a(n) = (4*n^3 +54*n^2 +212*n +153 -9*(-1)^n)/48.

G.f.: (2*x^3-4*x^2+3) / ((x-1)^4*(x+1)). - Colin Barker, Mar 04 2013

A064817 Maximal number of squares among the n-1 numbers p_i + p_{i+1}, 1 <= i <= n-1, where (p_1, ..., p_n) is any permutation of (1, ..., n).

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 16, 17, 18, 19, 20, 22, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68

Offset: 1

N. J. A. Sloane, Oct 23 2001

a(n) < n by definition, but if we counted the sum p_n + p_1, we could get a(n) = n for 32 <= n <= 49 (see A071984). - David Wasserman, Aug 20 2002
Can be formulated as a traveling salesman problem on a complete graph with node set {0, 1, ..., n} and edge cost -1 if i + j is a square, 0 otherwise. - Rob Pratt, Nov 07 2012
a(n) = n - 1 for 25 <= n <= 500, computed by solving corresponding TSP. - Rob Pratt, Nov 07 2012

			n=8: take 2,7,8,1,3,6,4,5 to get 5 squares: 2+7, 8+1, 1+3, 3+6, 4+5; a(8) = 5.
(1,8,9,7,2,14,11,5,4,12,13,3,6,10) gives 12 squares and no permutation of (1..14) gives more, so a(14)=12.

Bernardo Recamán Santos, Challenging Brainteasers, Sterling, NY, 2000, page 71, shows a(15) = 14 using 9,7,2,14,11,5,4,12,13,3,6,10,15,1,8.

Cf. A035106, A064764, A064796, A064797, A071983, A071984.

a[n_] := Which[n == 1, 0, n > 30, n - 1, True, tour = FindShortestTour[Range[n], DistanceFunction -> Function[{i, j}, If[IntegerQ[Sqrt[i + j]], -1, 0]]] // Last; cnt = 0; Do[If[IntegerQ[Sqrt[tour[[i]] + tour[[i + 1]]]], cnt++], {i, 1, n}]; cnt]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 69}] (* Jean-François Alcover, Nov 04 2016 *)

More terms from Vladeta Jovovic, Oct 23 2001
More terms from John W. Layman and Charles K. Layman (cklayman(AT)juno.com), Nov 07 2001
More terms from David Wasserman, Aug 20 2002
More terms from Rob Pratt, Nov 07 2012

A138585 The sequence is formed by concatenating subsequences S1, S2, ... each of finite length. S1 consists of the element 1. The n-th subsequence consist of numbers {(n/2)(n/2 - 1)+ 1, ..., (n/2)(n/2 + 1)} for n even, {((n-1)/2)^2, ..., (n-1)/2 * ((n-1)/2 + 2)} for n odd.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 8, 7, 8, 9, 10, 11, 12, 9, 10, 11, 12, 13, 14, 15, 13, 14, 15, 16, 17, 18, 19, 20, 16, 17, 18, 19, 20, 21, 22, 23, 24, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 31, 32, 33, 34, 35

Offset: 1

Ctibor O. Zizka, May 13 2008

A generalized Connell sequence.

Except for the first term the first element of each subsequence Sn (equivalently, each row of the triangle) gives A004652 (offset by 1), and the last element is A035106.

			S1: {1}
S2: {1,2}
S3: {1,2,3,}
S4: {3,4,5,6}
S5: {4,5,6,7,8}
S6: {7,8,9,10,11,12}, etc.
so concatenation of S1/S2/S3/S4/S5/S6/... gives:
1,1,2,1,2,3,3,4,5,6,4,5,6,7,8,7,8,9,10,11,12,...

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Douglas E. Iannucci and Donna Mills-Taylor. On Generalizing the Connell Sequence. Journal of Integer Sequences 2 (1999), Article 99.1.7.

Cf. A001614.

S := proc(n) local s: if(n=1)then s:=1: elif(n mod 2 = 0)then s:=(n/2)*(n/2 -1)+1: else s:=((n-1)/2)^2: fi: seq(k,k=s..s+n-1): end: seq(S(n),n=1..12); # Nathaniel Johnston, Oct 01 2011

Corrected and edited by D. S. McNeil, Dec 12 2010

A265436 a(n) is the least m (1 <= m <= n) such that the set of pairs (x, y) of distinct terms from [m, n] can be ordered in such a way that the corresponding sums (x+y) and products (x*y) are monotonic.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 35, 35, 36, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 47, 48, 48, 49, 50, 51, 52, 53, 54, 55

Offset: 1

Bill McEachen and Michel Marcus, Dec 09 2015

nonn

The pairs of distinct terms of [m,n] are first ordered according to their sums, then by their products.
This sequence seems related to both A183867 and A028387.
For A183867, let us define the sequence b(n) that gives the highest k such that a(k) = n. The data show that b(1)=4, b(2)=6, and the sequence b(n) begins 4, 6, 8, 9, 10, 12, 13, 15, 16, 17, 18, 20, ... and matches A183867(n+1) upwards.
Regarding A028387, its terms stem from the last (x,y) pair of each iteration, specifically its sum and product. From the examples provided below, for n=3 the last pair is (5,6) having sum 11. For n=5, the last pair is (9,20) having sum 29. These correspond to A028387(2) and A028387(4) respectively, and generally data from a(n) here produces A028387(n-1).
It appears that for n>5, the indices n where a(n)=a(n-1) are given by A035106(n). - Jean-François Alcover, Dec 20 2015

			For n=1, the only possible interval is [1,1], the set of distinct pairs is empty, so it satisfies the desired property, hence m=1 and a(1)=1.
For n=2, the candidate interval is [1,2], the set of distinct pairs is reduced to (1,2), which satisfies the order property hence m=1 and a(2)=1.
For n=3, the candidate interval is [1,2,3], with distinct pairs (1,2), (1,3), (2,3); and with corresponding sums (3,4,5) and products (2,3,6), that are monotonically ordered, hence m=1, so a(3)=1.
For n=5, the interval [1,5] fails to produce an ordering where both sums and products follow a monotonic order. But with m=2, here is a correct ordering: (5,6), (6,8), (7,10), (7,12), (8,15), (9,20); hence m=2 and a(5)=2.

QSYNT
Bill McEachen, Original Python program.

Cf. A028387, A035106, A060018, A183867.

pairs[m_, n_] := Flatten[Table[{x, y}, {x, m, n-1}, {y, x+1, n}], 1]; csum[ {x1_, y1_}, {x2_, y2_}] := x1+y1 <= x2+y2; cprod[{x1_, y1_}, {x2_, y2_}] := Which[x1 y1 < x2 y2, True, x1 y1 == x2 y2, x1+y1 <= x2+y2, True, False ]; a[1]=1; a[n_] := For[m=1, mJean-François Alcover, Dec 20 2015 *)

vpairs(n, m, nbp) = {v = vector(nbp); k = 1; for (i=m, n-1, for (j=i+1, n, v[k] = [i, j]; k++;)); v;}
vsums(v) = vector(#v, k, v[k][1] + v[k][2]);
vprods(v) = vector(#v, k, v[k][1] * v[k][2]);
cmpp(va, vb) = {sa = va[1]+va[2]; sb = vb[1]+vb[2]; if (sa > sb, return (1)); if (sa < sb, return (-1)); pa = va[1]*va[2]; pb = vb[1]*vb[2]; pa - pb;}
isok(n, m) = {nb = n-m+1; nbp = nb*(nb-1)/2; v = vpairs (n, m, nbp); perm = vecsort(v,cmpp,1); vs = vsums(v); vp = vprods(v); vss = vector(#vs, k, vs[perm[k]]); vps = vector(#vp, k, vp[perm[k]]); (vecsort(vps) == vps) && (vecsort(vss) == vss);}
one(n, m) = {ok = 0; while (!ok, if (! isok(n, m), m++, ok=1)); m;}
lista(nn) = {m = 1; for (n=1, nn, newm = one(n, m); print1(newm, ", "); m = newm;);}
\\ Michel Marcus, Dec 09 2015

def f1(X):
  x = X
  for y in range (1,X + 1):  # ie 1 thru X
    x = ((((((2 + y) * y) // (2 + x)) - 2) + x) // (2 + x)) + x    # floor division
  return x
def f0(X):
  return (f1(X) + 1) - X
for x in range(1000):
  print (f0(x))
# Bill McEachen, Jun 12 2024 (via the QSYNT link)

Conjecture (derived from the assumed relationship with A035106): for n>5, if sqrt(4n+1) is an odd integer or sqrt(n+1) is an integer, then a(n) = a (n-1), otherwise a(n) = a(n-1)+1. - Jean-François Alcover, Dec 21 2015

A289870 a(n) = n(n + 1) for n odd, otherwise a(n) = (n - 1)(n + 1).

Original entry on oeis.org

-1, 2, 3, 12, 15, 30, 35, 56, 63, 90, 99, 132, 143, 182, 195, 240, 255, 306, 323, 380, 399, 462, 483, 552, 575, 650, 675, 756, 783, 870, 899, 992, 1023, 1122, 1155, 1260, 1295, 1406, 1443, 1560, 1599, 1722, 1763, 1892, 1935, 2070, 2115, 2256, 2303, 2450, 2499

Offset: 0

Jean-François Alcover and Paul Curtz, Jul 14 2017

a(n) is a fifth-order linear recurrence whose main interest is that it is related to (at least) eight other sequences (see the formula section).

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

After -1, subsequence of A035106, A198442 and A214297.

Cf. A000466, A002378, A093178, A109613, A114285, A124356, A193356, A289296.

a[n_] := (n + 1)(n - 1 + Mod[n, 2]); Table[a[n], {n, 0, 50}]

a(n)=if(n%2, n, n-1)*(n+1) \\ Charles R Greathouse IV, Jul 14 2017

a(n) = (n + 1)*(n - 1 + (n mod 2)).
a(n) = n * A109613(n-1) for n>0.
a(n) = -A114285(n) * A109613(n).
a(n) = A002378(n) - A193356(n).
a(n) = A289296(-n).
a(n) = n^2 - (-1)^n * A093178(n).
a(2*k) = A000466(k).
G.f.: (1-3*x-3*x^2-3*x^3)/((-1+x)^3*(1+x)^2).

cp's OEIS Frontend

A163253 An interspersion: the order array of the odd-numbered columns of the double interspersion at A161179.

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A035104 First differences give (essentially) A028242.

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A064797 Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies lcm(p_i, p_{i+1}) >= m for some i, 1 <= i <= n, where p_{n+1} = p_1.

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A035107 First differences give (essentially) A028242.

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Magma

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A064817 Maximal number of squares among the n-1 numbers p_i + p_{i+1}, 1 <= i <= n-1, where (p_1, ..., p_n) is any permutation of (1, ..., n).

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A138585 The sequence is formed by concatenating subsequences S1, S2, ... each of finite length. S1 consists of the element 1. The n-th subsequence consist of numbers {(n/2)*(n/2 - 1)+ 1, ..., (n/2)*(n/2 + 1)} for n even, {((n-1)/2)^2, ..., (n-1)/2 * ((n-1)/2 + 2)} for n odd.

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A265436 a(n) is the least m (1 <= m <= n) such that the set of pairs (x, y) of distinct terms from [m, n] can be ordered in such a way that the corresponding sums (x+y) and products (x*y) are monotonic.

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A289870 a(n) = n*(n + 1) for n odd, otherwise a(n) = (n - 1)*(n + 1).

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A138585 The sequence is formed by concatenating subsequences S1, S2, ... each of finite length. S1 consists of the element 1. The n-th subsequence consist of numbers {(n/2)(n/2 - 1)+ 1, ..., (n/2)(n/2 + 1)} for n even, {((n-1)/2)^2, ..., (n-1)/2 * ((n-1)/2 + 2)} for n odd.

A289870 a(n) = n(n + 1) for n odd, otherwise a(n) = (n - 1)(n + 1).