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.

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A083480 Compress the triangular array A049597 by suppressing zero entries and reversing the order of each row.

Original entry on oeis.org

1, 2, 3, 4, 1, 5, 2, 6, 3, 2, 7, 4, 4, 8, 5, 6, 3, 9, 6, 8, 6, 1, 10, 7, 10, 9, 6, 11, 8, 12, 12, 11, 2, 12, 9, 14, 15, 16, 9, 2, 13, 10, 16, 18, 21, 16, 7, 14, 11, 18, 21, 26, 23, 18, 4, 15, 12, 20, 24, 31, 30, 29, 12, 3, 16, 13, 22, 27, 36, 37, 40, 27, 12, 1, 17, 14, 24, 30, 41, 44, 51
Offset: 1

Views

Author

Alford Arnold, Jun 08 2003

Keywords

Comments

Row sums => A000041. Diagonals are sums of Gaussian polynomials (which then sum to powers of two). The number of entries on each row is conjectured to conform to: 0 1 1 1 2 2 3 3 4 5 5 6 7 7 8 9 10 10 11 12 13 13 14 15 16 17 17 ... a sequence which stutters after values 0,1,2,4,6,9,12,16,...A002620.
Regarding the first element of the sequence as T(1,0), it appears that this is the number of partitions of n with k elements not in the first hook; i.e., with n - (max part size) - (number of parts) + 1 = k. If this is correct, we have T(n,0) = n and for k > 0, T(n,k) = sum_{j >= max(0,2k-n+2)} j * T(k,j). This is equivalent to T(n,k) = T(n-1,k) + sum_{j >= max(0,2k-n+2)} T(k,j) and thus to T(n,k) = 2* T(n-1,k) - T(n-2,k) + T(k,2k-n+2) [taking T(n,k) = 0 if k < 0]. It also implies the correctness of the conjecture about the row lengths. - Franklin T. Adams-Watters, May 27 2008

Examples

			The table begins:
1
2
3
4 1
5 2
6 3 2
7 4 4
8 5 6 3
9 6 8 6 1
...
		

Crossrefs

Programs

  • Maple
    a:=n->sort(simplify(sum(product((1-q^i),i=n-r+1..n)/product((1-q^j),j=1..r), r=0..n))):T := proc(n,k) if k=n then n+1 elif k>n then 0 else coeff(a(k),q^(n-k)) fi end: b:=proc(n,k) if T(n,n-k)>0 then T(n,n-k) else fi end:seq(seq(b(n,k),k=0..n+1),n=0..20); # Emeric Deutsch, May 15 2004
  • Mathematica
    a[n_] := Sum[Product[1-q^i, {i, n-r+1, n}]/Product[1-q^j, {j, 1, r}], {r, 0, n}] // Simplify; T [n_, k_] := Which[k == n, n+1, k>n, 0, True, Coefficient[a[k], q^(n - k)]]; Table[Table[T[n, k], {k, n, 0, -1}] // DeleteCases[#, 0]&, {n, 0,  21}] // Flatten (* Jean-François Alcover, Feb 19 2015, after Maple *)

Extensions

More terms from Emeric Deutsch, May 15 2004

A105552 Irregular triangle T(n,k) read down columns: the number of compositions c of n with largest_part(c)+length(c)=k+1 in row n, column k.

Original entry on oeis.org

1, 2, 4, 1, 7, 5, 2, 11, 14, 12, 5, 1, 16, 30, 39, 32, 18, 7, 2, 22, 55, 95, 113, 101, 71, 41, 18, 6, 1, 29, 91, 195, 299, 357, 350, 292, 207, 126, 64, 27, 9, 2, 37, 140, 357, 664, 978, 1204, 1283, 1198, 992, 731, 482, 284, 148, 66, 25, 7, 1, 46, 204, 602, 1309, 2274, 3329, 4253
Offset: 1

Views

Author

Alford Arnold, May 03 2005

Keywords

Comments

For each of the A000041(n) partitions of n, one can assign a weight to the partition which counts the permutations of that partition, given by the multinomial coefficient derived from the frequency representation of the parts.
An equivalent representation is given by writing down all compositions of n.
The entries count those partitions multiplied by their weights (=compositions) of n where the sum of the largest addend plus number of parts equals k+1. Only nonzero counts are entered into the sequence.
Each entry can also be interpreted as counting a subset of numbers in A055932, because there is a 1-to-1 correspondence between their prime signature and ordered partitions.
Each diagonal of T(n,k) can be decomposed into p(n) sequences. For example,
A086602 = 2 12 39 95 195 ... is the sum of
A000330 = 1 5 14 30 55 ... plus
A001296 = 1 7 25 65 140 ...
The main diagonal and subdiagonals in order of appearance are A000124, A000330, A086602, A089574, A107600, A107601, A109125, ...

Examples

			The row n=7 starts from the partitions (weights in parentheses) 7 (1), 6+1 (2), 5+2 (2), 4+3 (2), 5+1+1 (3), 4+2+1 (6=3!/1!/1!/1!), 3+3+1 (3), 3+2+2 (3), 4+1+1+1 (4=4!/1!/3!), 3+2+1+1 (12 = 4!/1!/1!/2!), 2+2+2+1 (4), 3+1+1+1+1+1 (5), 2+2+1+1+1 (10=5!/2!/3!), 2+1+1+1+1 (6), 1+1+1+1+1+1 (1).
Then T(7,7) = 1+2+3+4+5+6+1 = 22 is the sum of the weights of partitions with largest part 7 and length 1, largest part 6 and length 2,... largest part 1 and length 7.
T(7,6) = 2+6+12+10 = 30 is the sum of the weights of the partitions with largest part 6 and length 1, largest part 5 and length 2, ..., largest part 1 and length 6.
T(7,5) = 2+3+3+4 = 12 collects all the partitions with largest part 5 and length 1 down to largest part 1 and length 5.
The array has A033638(k) nonzero entries per column, starting at n=1 as :
1
..2
....4
....1..7
.......5..11
.......2..14..16
..........12..30..22
...........5..39..55..29
...........1..32..95..91..37
..............18.113.195.140
...............7.101.299.357
...............2
		

Crossrefs

Cf. A047969, A047970, A055932, A057335, A083480, A083906, A089349, A033638, A086602 (subdiagonal), A089574 (subdiagonal).

Programs

  • Maple
    A033638 := proc(n) ( (7+(-1)^n)/2 + n^2 )/4 ; end proc:
    freq := proc(L,n) local a,p; a := 0 ; for p in L do if p = n then a := a+1 ; end if; end do: a ; end proc:
    M3 := proc(L) local a,i; a := factorial(nops(L)) ; for i in convert(L,set) do a := a/factorial(freq(L,i)) ; end do: a ; end proc:
    A105552 := proc(n,k) local p,a,l ; a := 0 ; for p in combinat[partition](n) do if max(op(p)) + nops(p) = k+1 then a := a+ M3(p); end if; end do ; a ; end proc:
    for k from 1 to 15 do for n from k to k+A033638(k)+1 do T := A105552(n,k) ; if T >0 then printf("%d,", A105552(n,k)) ; end if; end do: printf("\n") ; end do: # R. J. Mathar, Jun 26 2010
    # second Maple program:
    b:= proc(n, k, p) option remember; `if`(n=0 and k=0, 1,
         `if`(k<1, 0, add(b(n-j, k-1-max(p, j)+p, max(p, j)), j=1..n)))
        end:
    T:= k-> seq(b(n, k+1, 0), n=k..k+floor((k-1)^2/4)):
    seq(T(k), k=1..10);  # Alois P. Heinz, Jul 24 2013
  • Mathematica
    b[n_, k_, p_] := b[n, k, p] = If[n == 0 && k == 0, 1, If[k < 1, 0, Sum[b[n - j, k - 1 - Max[p, j] + p, Max[p, j]], {j, 1, n}]]]; T[k_] := Table[b[n, k + 1, 0], {n, k, k + Floor[(k - 1)^2/4]}]; Table[T[k], {k, 1, 10}] // Flatten (* Jean-François Alcover, Jan 08 2016, after Alois P. Heinz *)

Formula

Row sums: Sum_{k=0..n} T(n,k) = 2^(n-1).
Column sums: Sum_{n>=k} T(n,k) = A047970(n).

Extensions

Definition clarified by R. J. Mathar, Jun 26 2010

A112970 A generalized Stern sequence.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 2, 2, 1, 5, 3, 3, 2, 5, 2, 3, 1, 6, 4, 3, 2, 6, 2, 3, 1, 7, 5, 4, 3, 8, 3, 5, 2, 8, 5, 4, 2, 8, 3, 3, 1, 9, 6, 5, 4, 9, 3, 6, 2, 9, 6, 4, 2, 9, 3, 3, 1, 10, 7, 6, 5, 11, 4, 8, 3, 12, 8, 6, 3, 13, 5, 5, 2, 13, 8, 7, 5, 12, 4, 7, 2, 12, 8, 5, 3, 11, 3, 4, 1, 12, 9, 7, 6
Offset: 0

Views

Author

Paul Barry, Oct 07 2005

Keywords

Comments

Conjectures: a(2^n)=a(2^(n+1)+1)=A033638(n); a(2^n-1)=a(3*2^n-1)=1.
The Gi1 and Gi2 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence. The Gi1 and Gi2 sums can also be interpreted as (i + 4*j = n) and (4*i + j = n) sums, see the Northshield reference. Some A112970(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs. - Johannes W. Meijer, Jun 05 2011

Crossrefs

Cf. A120562 (Northshield).
Cf. A033638 (p=0), A000012 (p=1), A004526 (p=2, p=3, p=5, p=9, p=17), A002620 (p=4, p=7, p=13, p=25), A000027 (p=6, p=11, p=21), A004116 (p=8, p=15, p=29), A035106 (p=10, p=19), A024206 (p=14, p=27), A007494 (p=18), A014616 (p=22), A179207 (p=26). - Johannes W. Meijer, Jun 05 2011

Programs

  • Maple
    A112970:=proc(n) option remember; if n <0 then A112970(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A112970(n/2) + A112970((n/2)-2) else A112970((n-1)/2); fi; end: seq(A112970(n),n=0..99); # Johannes W. Meijer, Jun 05 2011
  • Mathematica
    a[n_] := a[n] = Which[n<0, 0, n==0 || n==1, 1, Mod[n, 2]==0, a[n/2] + a[n/2-2], True, a[(n-1)/2]];
    Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Aug 02 2022 *)

Formula

a(n) = Sum_{k=0..n} mod(sum{j=0..n, (-1)^(n-k)*C(j, n-j)*C(k, j-k)}, 2).
From Johannes W. Meijer, Jun 05 2011: (Start)
a(2*n+1) = a(n) and a(2*n) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.
G.f.: Product_{n>=0} (1 + x^(2^n) + x^(4*2^n)). (End)
G.f. A(x) satisfies: A(x) = (1 + x + x^4) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019

A215940 Difference between the n-th and the first (identity) permutation of (0,...,m-1), interpreted as a decimal number, divided by 9 (for any m for which 10! >= m! >= n).

Original entry on oeis.org

0, 1, 10, 12, 21, 22, 100, 101, 120, 123, 131, 133, 210, 212, 220, 223, 242, 243, 321, 322, 331, 333, 342, 343, 1000, 1001, 1010, 1012, 1021, 1022, 1200, 1201, 1230, 1234, 1241, 1244, 1310, 1312, 1330, 1334, 1352, 1354, 1421, 1422, 1441, 1444, 1452, 1454, 2100
Offset: 1

Views

Author

R. J. Cano, Sep 21 2012

Keywords

Comments

Original definition: "Quotients of the polynomial remainder theorem for Diophantine equations among permutations."
The set built from the first terms of this sequence {0, 1, 10, 12, 21, 22, ....} contains the general solutions for a class of Diophantine linear equations among permutations, if one takes into account the unlimited number of distinct bases where these may be read.
From M. F. Hasler, Jan 12 2013, edited by R. J. Cano, May 08 2017: (Start)
Let P be the sequence of permutations of (0,...,m-1) interpreted as decimal numbers, P(n) = Sum_{i=1..m} 10^(m-i)*s(i) where s=(s(1),...,s(m)) is the n-th permutation (in lexicographical order), n <= m!. Then the difference P(n)-P(1) is independent of the choice of m, and divisible by 9. (Since 10 == 1 (mod 9), the numbers P(n) are all congruent (mod 9) to the sum s(1)+...+s(m).) This yields well-defined terms a(n)=(P(n)-P(1))/9.
For n>10!, P(n) will no longer be the concatenation of the "digits" (some of which will exceed 9). The pattern present in the decimal representation of the first terms will also be lost since there will be digits as large as d+1.
Note that the same a(n) is obtained independently of the chosen base b, provided that (i) 10 and 9 in the above are replaced with b and b-1, (ii) the result is (and can be) written in base b. (This implies the restriction to terms which can be written using the digits 0-9 to which the OEIS is limited.) See EXAMPLES for an illustration. (End)
We have P(n)-P(1)=a(n)*g(n), with g(n) = 9 = 10-1. Considering this and P(n) as polynomials in x=10, one can see an analogy with the polynomial remainder theorem. [Given as "formula" by R. J. Cano, rephrased by M. F. Hasler, Jan 12 2013]
Contribution by R. J. Cano, Feb 09 2013, (Start)
The maximum of the first m! terms of this sequence is given in base R by the explicit formula (please see A211869): max(m,R)=Sum_{k=1..m} k*(m-k)*R^(m-k-1);
If the first m! terms of this sequence were computed reading the permutations in base A033638(m), dividing their differences by A033638(m)-1, the resulting quotients would be written in the same way (with the same digits) in every base > A033638(m).
(End)
From R. J. Cano, Apr 29 2016, (Start)
Although in the sequence name it reads: "permutation of (0,...,m-1)", the most general statement that could replace it is: "permutation of any m-tuple of integers all of them in arithmetic progression", obtaining a multiple of this sequence, lambda*a(n), where lambda is the common difference for the progression. It works in such way because only the differences is what matters here.
Given x>1 and k>=0, if a polynomial G(x) of degree k is divided by x-1 then the remainder will be the sum of all the coefficients in G. Let us consider the case in which those coefficients are the differences among the letters ("digits") of two permutations for the same set of letters (0..x-1): The sum of all those differences must vanish. This explains how the difference between two of such permutations expressed in base x is 0 mod x-1, particularly why differences for pairs of permutations are divisible by 9.
Another way of introducing this sequence takes advantage of the fact that for n>1, n! is even. Consider for n>1 to obtain only the first n! terms. This can be done by subtracting the last permutation from the first, the penultimate permutation from the second, and so on by following the pattern (P(k)-P(n!-k+1))/9 with 1<=k<=n!. Such procedure generates an antisymmetric sequence f(k) from which a(k)=(f(k)+f(n!))/2. This partially explains why A217626 is symmetric. Also a base-independent treatment is possible using linear algebra: Column vectors and the strictly lower triangular matrix instead of the division by (r-1) where r is the base (and r=10 here for this sequence). This approach leads one to conclude that terms in this sequence are the differences between pairs of vectors made from the first n-1 partial sums of letters ("digits") taken from permutations for n consecutive letters, when components in these vectors are viewed as coefficients for a power series in r=10.
(End)

Examples

			From _M. F. Hasler_, Jan 12 2013, edited by _R. J. Cano_, May 09 2017: (Start)
The permutations of {0,1,2}, read as numbers, are {12, 21, 102, 120, 201, 210}. Subtracting the first one (12) from each of these numbers yields the differences {0, 9, 90, 108, 189, 198}. These are all multiples of 9, see comment and links. Dividing the differences by 9 yields {0, 1, 10, 12, 21, 22}, which are by definition the first six terms of this sequence.
Using all permutations of 0123 would yield 4!=24 terms, where the first 6 would be identical to those above. For n>10! we must consider permutations of (0,...,m-1) with m>10. These are no longer valid digits in base 10, and the numbers P(n) as defined in the comment are no longer equal to the concatenation. However, the first 10! terms obtained as (P(n)-P(1))/9, are still the same as for m=10;
In order to illustrate that the result is independent of the base chosen to make the calculation, let us consider permutations of 012 in base 3. The 3rd resp. 5th term ((102-012)/9=10 resp. (201-012)/9=21) would be ((1-0)*3^2+(0-1)*3+(2-2)*1)/(3-1) = 3 = 10[3], resp. ((2-0)*3^2+(0-1)*3+(1-2)*1)/(3-1) = 7 = 21[3]. The same terms would also be "10" and "21" if the calculation were made in base b=11. In that base, with digit "A" having the value b-1, we have (1023456789A - 0123456789A)/A = 0A000000000[11], (2013456789A - 0123456789A)/A = 02100000000[11], and (0A123456789 - 0123456789A)/A = 0A987654321[11] (the analog of (40123[5]-01234[5])/4[5] = 04321[5]). (End)
		

Crossrefs

Programs

  • C
    // See links.
    
  • Maple
    N:= 100: # to get a(1)..a(N)
    for M from 3 while M! <= N do od:
    p0:= [$1..M]: p:= p0: A[1]:= 0:
    for n from 2 to N do
      p:= combinat:-nextperm(p);
      d:= p - p0;
      A[n]:= add(10^(i-1)*d[-i],i=1..M)/9;
    od:
    seq(A[i],i=1..N); # Robert Israel, Apr 19 2017
  • Mathematica
    maxm = 5; Table[dd = FromDigits /@ Permutations[Range[m]]; (Drop[dd, If[m == 1, 0, (m - 1)!]] - First[dd])/9, {m, 1, maxm}] // Flatten (* Jean-François Alcover, Apr 25 2013 *)
  • PARI
    A215940(n)=for(k=2, n+1, k!M. F. Hasler, Jan 12 2013
    
  • PARI
    first_n_factorial_terms(n)={my(u=n!);my(x=numtoperm(n,0),y,z=vector(u),i:small);i=0;for(j=0,u-1,y=numtoperm(n,j)-x;z[i++]=fromdigits(vector(#x-1,k,vecsum(y[1..k]))));z} \\ R. J. Cano, Apr 19 2017

Formula

a(n) = Sum_{k=1..n-1} A217626(k).
a(n) = (A050289(n)-A050289(1))/9, for n <= 9!. - M. F. Hasler, Jan 12 2013

Extensions

Edited by M. F. Hasler, Jan 12 2013
Minor edits by N. J. A. Sloane, Feb 19 2013

A126256 Number of distinct terms in rows 0 through n of Pascal's triangle.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 9, 12, 16, 20, 24, 29, 35, 41, 48, 53, 60, 68, 77, 86, 95, 103, 114, 125, 137, 149, 162, 175, 188, 202, 217, 232, 248, 264, 281, 297, 314, 332, 351, 370, 390, 410, 431, 452, 474, 495, 518, 541, 565, 589, 614, 639, 665, 691, 718, 744, 770, 798
Offset: 0

Views

Author

Nick Hobson, Dec 24 2006

Keywords

Comments

An easy upper bound is 1 + floor(n^2/4) = A033638(n).
First differences are in A126257.

Examples

			There are 9 distinct terms in rows 0 through 6 of Pascal's triangle (1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1); hence a(6)=9.
		

Crossrefs

Programs

  • Haskell
    -- import Data.List.Ordered (insertSet)
    a126256 n = a126256_list !! n
    a126256_list = f a007318_tabl [] where
       f (xs:xss) zs = g xs zs where
         g []     ys = length ys : f xss ys
         g (x:xs) ys = g xs (insertSet x ys)
    -- Reinhard Zumkeller, May 26 2015, Nov 09 2011
    
  • Maple
    seq(nops(`union`(seq({seq(binomial(n,k),k=0..n)},n=0..m))),m=0..57); # Emeric Deutsch, Aug 26 2007
  • Mathematica
    Table[Length[Union[Flatten[Table[Binomial[n,k],{n,0,x},{k,0,n}]]]],{x,0,60}] (* Harvey P. Dale, Sep 10 2022 *)
  • PARI
    lim=57; z=listcreate(1+lim^2\4); for(n = 0, lim, for(r=1, n\2, s=Str(binomial(n, r)); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); print1(1+#z, ", "))
    
  • Python
    def A126256(n):
        s, c = (1,), {1}
        for i in range(n):
            s = (1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + (1,)
            c.update(set(s))
        return len(c) # Chai Wah Wu, Oct 17 2023

A278354 Number of neighbors of each new term in a square spiral.

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 2, 4, 3, 2, 4, 3, 2, 4, 4, 3, 2, 4, 4, 3, 2, 4, 4, 4, 3, 2, 4, 4, 4, 3, 2, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4
Offset: 1

Views

Author

Omar E. Pol, Nov 19 2016

Keywords

Comments

Here the "neighbors" of a(n) are defined to be the adjacent elements to a(n) in the same row, column or diagonals, that are present in the spiral when a(n) is the new element of the sequence in progress.
For the same idea but for a right triangle see A278317; for an isosceles triangle see A275015; for a square array see A278290; and for a hexagonal spiral see A047931.

Examples

			Illustration of initial terms as a spiral (n = 1..169):
.
.     2 - 3 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 2
.     |                                               |
.     4   2 - 3 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 2   3
.     |   |                                       |   |
.     4   4   2 - 3 - 4 - 4 - 4 - 4 - 4 - 4 - 2   3   4
.     |   |   |                               |   |   |
.     4   4   4   2 - 3 - 4 - 4 - 4 - 4 - 2   3   4   4
.     |   |   |   |                       |   |   |   |
.     4   4   4   4   2 - 3 - 4 - 4 - 2   3   4   4   4
.     |   |   |   |   |               |   |   |   |   |
.     4   4   4   4   4   2 - 3 - 2   3   4   4   4   4
.     |   |   |   |   |   |       |   |   |   |   |   |
.     4   4   4   4   4   3   0 - 1   4   4   4   4   4
.     |   |   |   |   |   |           |   |   |   |   |
.     4   4   4   4   3   2 - 4 - 3 - 2   4   4   4   4
.     |   |   |   |   |                   |   |   |   |
.     4   4   4   3   2 - 4 - 4 - 4 - 3 - 2   4   4   4
.     |   |   |   |                           |   |   |
.     4   4   3   2 - 4 - 4 - 4 - 4 - 4 - 3 - 2   4   4
.     |   |   |                                   |   |
.     4   3   2 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 3 - 2   4
.     |   |                                           |
.     3   2 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 3 - 2
.     |
.     2 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 4 - 3
.
		

Crossrefs

Programs

  • Maple
    0,1,seq(op([2,4$floor(i/2),3]),i=0..30); # Robert Israel, Nov 22 2016

Formula

From Robert Israel, Nov 22 2016: (Start)
a(n) = 3 if n>=4 is in A002620.
a(n) = 2 if n>=2 is in A033638.
Otherwise, a(n) = 4 if n > 2. (End)

A056537 Mapping from the column-by-column reading to the half-antidiagonal reading of the triangular tables. Inverse of A056536.

Original entry on oeis.org

1, 2, 4, 3, 6, 9, 5, 8, 12, 16, 7, 11, 15, 20, 25, 10, 14, 19, 24, 30, 36, 13, 18, 23, 29, 35, 42, 49, 17, 22, 28, 34, 41, 48, 56, 64, 21, 27, 33, 40, 47, 55, 63, 72, 81, 26, 32, 39, 46, 54, 62, 71, 80, 90, 100, 31, 38, 45, 53, 61, 70, 79, 89, 99, 110, 121, 37, 44, 52, 60, 69
Offset: 1

Views

Author

Antti Karttunen, Jun 20 2000

Keywords

Comments

Moves triangular numbers (A000217) to squares (A000290), i.e., A056537(A000217(i)) = A000290(i) for i >= 1.
As a square array, this is the dispersion of the complement of the squares; see A082152. - Clark Kimberling, Apr 05 2003

Examples

			As a square array, a northwest corner:
1 ... 2 ... 3 ... 5 ... 7 ... 10
4 ... 6 ... 8 ... 11 .. 14 .. 18
9 ... 12 .. 15 .. 19 .. 23 .. 28
16 .. 20 .. 24 .. 29 .. 34 .. 40
25 .. 30 .. 35 .. 41 .. 47 .. 54
36 .. 42 .. 48 .. 55 .. 62 .. 70
49 .. 56 .. 63 .. 71 .. 79 .. 88
64 .. 72 .. 80 .. 89 .. 98 .. 108
- _Clark Kimberling_, Aug 08 2013
		

Crossrefs

Cf. A185787 (dispersion of complement of triangular numbers).
Cf. A082152 (dispersion of complement of pentagonal numbers).

Programs

  • Maple
    # using Maple procedure nthmember given in A054426:
    [seq(nthmember(j, A056536), j=1..105)];
  • Mathematica
    (* Program generates the dispersion array T of the increasing sequence f[n] *)
    r=40; r1=12; c=40; c1=12; f[n_] := n+Floor[1/2+Sqrt[n]] (* complement of column 1 *); mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]; rows = {NestList[f, 1, c]}; Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}]; t[i_, j_] := rows[[i, j]]; TableForm[Table[t[i, j], {i, 1, r1}, {j, 1, c1}]]  (* A056537 array *)
    Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A056537 sequence *)
    (* Clark Kimberling, Jun 06 2011 *)

Formula

Triangle T(n, k), 1<=k<=n, read by rows, defined by: T(n, k) = 0 for nA002620(n-k+1) + k*n + k - n if n>=k. T(n, n) = n^2; T(n, 1) = 1 + A002620(n) = A033638(n). - Philippe Deléham, Feb 16 2004
Square: t(n,k) = (n-1)(n+k) + k^2/4 + (1/8)(7+(-1)^k). - Clark Kimberling, Aug 08 2013

A352838 Irregular triangle read by rows: T(n, k) is the number of 2n-step closed walks on the square lattice having algebraic area k; n >= 0, 0 <= k <= floor(n^2/4).

Original entry on oeis.org

1, 4, 28, 4, 232, 72, 12, 2156, 1008, 308, 48, 8, 21944, 13160, 5540, 1560, 420, 80, 20, 240280, 168780, 87192, 33628, 11964, 3636, 1200, 264, 72, 12, 2787320, 2168544, 1291220, 610232, 262612, 101976, 40376, 13720, 4900, 1512, 420, 112, 28
Offset: 0

Views

Author

Andrei Zabolotskii, Apr 05 2022

Keywords

Comments

Rows can be extended to negative k with T(n, -k) = T(n, k). Sums of such extended rows give A002894.
T(n, k) is the number of words of length 2n equal to z^k in the Heisenberg group, presented as , where z=[x,y]. In particular, T(n, 0) = A307468(n).

Examples

			The table begins:
       1
       4
      28,      4
     232,     72,    12
    2156,   1008,   308,    48,     8
   21944,  13160,  5540,  1560,   420,   80,   20
  240280, 168780, 87192, 33628, 11964, 3636, 1200, 264, 72, 12
     ...
T(2, 0) = 28: the 4-step walks enclosing algebraic area 0 include 16 walks of the form "some two steps, then two steps right back" and 12 walks of the form "some step, step back, a different step, step back".
T(2, 1) = 4: the 4-step walks enclosing algebraic area 1 are the walks around each of the 4 squares touching the origin in the positive direction; cf. A334756(2, 1) = 8, which also counts walks around these squares in the negative direction.
		

Crossrefs

Row lengths are A033638 = A002620 + 1.
Row n ends with 4 * A026741(n) for n > 0.
Row 16 is A178106.
A334756 counts self-avoiding walks only.

Programs

  • Mathematica
    z[0, 0, 0, 0] = 1;
    z[-1, ] = z[, -1, _] = z[, , -1, ] = z[, , , -1] = 0;
    z[m1_, m2_, l1_, l2_] := z[m1, m2, l1, l2] = Expand[z[m1, m2, l1 - 1, l2] + z[m1, m2, l1, l2 - 1] + q^(l2 - l1) z[m1 - 1, m2, l1, l2] + q^(l1 - l2) z[m1, m2 - 1, l1, l2]];
    zN[n_] := Sum[z[m, m, n/2 - m, n/2 - m], {m, 0, n/2}];
    walks[n_] := Module[{gf = zN[2 n], e}, e = Exponent[gf, q, Max]; CoefficientList[gf q^e, q][[e + 1 ;;]]];
    Table[walks[n], {n, 0, 8}]

A383185 Number of the square visited by a king moving on a spirally numbered board always to the lowest available unvisited square, when a wall delimiting the spiral must be crossed on each move.

Original entry on oeis.org

0, 3, 13, 2, 10, 1, 7, 21, 6, 18, 4, 14, 32, 12, 28, 11, 27, 9, 23, 8, 22, 44, 20, 40, 19, 5, 17, 37, 16, 34, 15, 33, 59, 31, 57, 30, 54, 29, 53, 85, 51, 25, 47, 24, 46, 76, 45, 75, 43, 73, 42, 70, 41, 69, 39, 67, 38, 66, 36, 62, 35, 61, 95, 60, 94, 58, 92, 56, 88, 55, 87, 127, 86, 52, 26
Offset: 0

Views

Author

M. F. Hasler, May 12 2025

Keywords

Comments

The board is numbered following a square spiral starting with 0 at the origin (where the king is at n = 0) and delimited by a wall that must be crossed on each move:
.
16 15 14 13 12 | .
,-----------. | .
17 | 4 3 2 |11 | .
| ,--- | | .
18 | 5 | 0 1 |10 | .
| '-------' | .
19 | 6 7 8 9 | .
`---------------' .
20 21 22 23 24 25
.
A line drawn from the center of the starting square to the center of the ending square must pass through a wall on each move. A move that would just touch a wall without passing through the wall (e.g., 0 to 2) is not permissible. Equivalently, the king can't move from a square labeled k to a square labeled k +- 1 or k +- 2, i.e., |a(n)-a(n+1)| > 2 for all n.
This sequence is a permutation of the nonnegative integers, see A383186 for the inverse permutation. The king's walk indeed fills the 2D grid with an initial segment S0 of 24 moves, followed by rings R(r), r >= 1, which consist of three shells S1(r), S2(r) and S3(r), each of which corresponds to a tour around the center. Each ring R(r) starts with the move number n = 48 r^2 - 16 r + 1 = (25, 147, 365, ...) to the square at position P(r) = (2-3r, 3r-3) = ((-1,0), (-4,3), (-7,6), ...), and contains a perfectly well defined sequence of 96 r + 26 grid points following a precise sequence of pattern given in full detail on the wiki page provided in the links section.

Examples

			For n = 1, a(1) = 3 because moving from 0 to 1 or 2 does not pass through a wall.
		

Crossrefs

Cf. A375925 (the same with indices and numbers of squares starting at 1).
Cf. A383186 (inverse permutation).
Cf. A316328 (knight's path), A033638, A316667 (trapped knight), A336038 (trapped king), A335856 (trapped king preferably moving to prime numbers).

Programs

  • Python
    def square_number(z): return int(4*y**2-y-x if (y := z.imag) >= abs(x := z.real)
        else 4*x**2-x-y if -x>=abs(y) else (4*y-3)*y+x if -y>=abs(x) else (4*x-3)*x+y)
    def A383185(n):
        if not hasattr(A:=A383185, 'terms'): A.terms=[0]; A.pos=0; A.path=[0]
        while len(A.terms) <= n:
            s,d = min((s,d) for d in (1, 1+1j, 1j, 1j-1, -1, -1-1j, -1j, 1-1j) if
                abs((s:=square_number(A.pos+d))-A.terms[-1]) > 2 and s not in A.terms)
            A.terms.append(s); A.pos += d; A.path.append(A.pos)
        return A.terms[n]
    import matplotlib.pyplot as plt # this and below to plot the trajectory
    plt.plot([z.real for z in A383185.path], [z.imag for z in A383185.path])
    plt.show()

A193416 Minimum surface area of polycubes with volume n.

Original entry on oeis.org

6, 10, 14, 16, 20, 22, 24, 24, 28, 30, 32, 32, 36, 38, 40, 40, 42, 42, 46, 48, 50, 50, 52, 52, 54, 54, 54, 58, 60, 62, 62, 64, 64, 66, 66, 66, 70, 72, 74, 74, 76, 76, 78, 78, 78, 80, 80, 80, 84, 86, 88, 88, 90, 90, 92, 92, 92, 94, 94, 94, 96, 96, 96, 96, 100
Offset: 1

Views

Author

Juan Barajas Martin, Jul 25 2011

Keywords

Comments

First differences are 0, 2, or 4. - Charles R Greathouse IV, Aug 25 2011
From Juan Barajas Martin, Aug 27 2011 - Sep 03 2011, Sep 26 2001: (Start)
Initial cubic shape polycube edge = e; volume = e^3; minimum surface area = 6e^2.
Target cubic shape polycube edge = e+1; volume = (e+1)^3; minimum surface area = 6(e+1)^2.
The target polycube is achieved by the successive addition of 3 orthogonal layers of unit cubes to the faces of the initial polycube. The number of unit cubes added in each successive layer take values e^2, e*(e+1) and (e+1)^2, with every layer fully completed before starting the following layer.
This is so because the first unit cube added on every layer will add a surface area of 4 to the previous polycube, which is greater than keeping the new unit cube in the current layer, where we ensure that area will increase by 2 at maximum.
Every layer is constructed as an Ulam spiral, starting at the center and until the completion of the layer.
Except for the first cube placed on every layer which will increase the surface area of the previous polycube by 4, we can count 2 surface area increments for the unit cubes placed at the positions given by the function of the Quarter-squares + 1 sequence (i.e., A033638 = A002620 + 1, which set the locations of right angle turns in Ulam square spiral).
All other positions in the Ulam spiral will increase no area to the previous polycube.
An infinite sequence of polycubes with minimal surface area is approached between pairs of successive polycubes with cubic shape and edge values (e) and (e+1), having respectively polycube volumes e^3 and (e+1)^3 and minimum surface areas 6e^2 and 6(e+1)^2.
(End)

Examples

			The unique polycube of volume 1 is a cube with surface area 6, so a(1) = 6. There are eight polycubes of volume 4, of which seven have surface area 18 and one has surface area 16, so a(4) = 16. - _Charles R Greathouse IV_, Aug 25 2011
		

Crossrefs

Cf. A033638 (A002620 + 1) identify the unit cubes which will increase the minimum area by 2 (locations of right angle turns in Ulam square spiral), A007818.

Programs

  • Mathematica
    vals=100;
    az[n_]:=Floor[(n-1)^(1/3)];
    kz[n_]:=n-az[n]^3;
    av[n_]:=6*az[n]^2;
    bv[n_]:=If[n==1,4,If[kz[n]>az[n]^2+(az[n]+1)*az[n],12,If[kz[n]>az[n]^2,8,4]]];
    pz[n_]:=If[kz[n]1,2*(c1[n]+c2[n]+c3[n]),2];
    smin[n_]:=av[n]+bv[n]+cv[n];
    Table[smin[n], {n,vals}] (* Juan Barajas Martin, Sep 01 2011 *)

Formula

a(n^3) = 6n^2, a(n) ~ 6n^(2/3). - Charles R Greathouse IV, Aug 25 2011
From Juan Barajas Martin, Aug 28 2011: (Start)
The following formula is derived from the Mathematica program below:
smin[n]={6 Floor[(-1+n)^(1/3)]^2+If[n==1,4,If[kz[n]>az[n]^2+az[n] (az[n]+1),12,If[kz[n]>az[n]^2,8,4]]]+If[n>1,2 (c1[n]+c2[n]+c3[n]),2]}
az[n_]:Floor[Power[n-1, (3)^-1]]
kz[n_]:=n-az(n)^3
c1-c3: number of unit cubes increasing the surface area by 2 in every layer (see Comments above). (End)
a(n) = 6*n - 2*A007818(n). - Mohammed Yaseen, Aug 08 2021
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