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 A237271 Number of parts in the symmetric representation of sigma(n). 148
 1, 1, 2, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 2, 4, 1, 2, 1, 2, 1, 4, 2, 3, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 2, 2, 1, 3, 3, 4, 2, 2, 1, 4, 1, 4, 2, 2, 1, 2, 2, 5, 1, 4, 1, 2, 2, 4, 3, 2, 1, 2, 2, 4, 2, 3, 2, 2, 1, 5, 2, 2, 1, 4, 2, 4, 1, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS The diagram of the symmetry of sigma has been obtained according to the following way: A196020 --> A236104 --> A235791 --> A237591 --> A237593. For more information see A237270. a(n) is also the number of terraces at n-th level (starting from the top) of the step pyramid described in A245092. - Omar E. Pol, Apr 20 2016 a(n) is also the number of subparts in the first layer of the symmetric representation of sigma(n). For the definion of "subpart" see A279387. - Omar E. Pol, Dec 08 2016 Note that the number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. (See the second example). - Omar E. Pol, Dec 20 2016 From Hartmut F. W. Hoft, Dec 26 2016: (Start) Using odd prime number 3, observe that the 1's in the 3^k-th row of the irregular triangle of A237048 are at index positions      3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...   the last being 2*3^((k-1)/2) when k is odd and 3^(k/2) when k is even. Since odd and even index positions alternate, each pair (3^i, 2*3^i) specifies one part in the symmetric representation with a center part present when k is even. A straightforward count establishes that the symmetric representation of 3^k, k>=0, has k+1 parts. Since this argument is valid for any odd prime, every positive integer occurs infinitely many times in the sequence. (End) LINKS Michel Marcus, Table of n, a(n) for n = 1..5000 FORMULA a(p^k) = k + 1, where p is an odd prime and k >= 0. - Hartmut F. W. Hoft, Dec 26 2016 EXAMPLE Illustration of initial terms (n = 1..12): --------------------------------------------------------- n   A000203  A237270    a(n)            Diagram --------------------------------------------------------- .                               _ _ _ _ _ _ _ _ _ _ _ _ 1       1      1         1     |_| | | | | | | | | | | | 2       3      3         1     |_ _|_| | | | | | | | | | 3       4      2+2       2     |_ _|  _|_| | | | | | | | 4       7      7         1     |_ _ _|    _|_| | | | | | 5       6      3+3       2     |_ _ _|  _|  _ _|_| | | | 6      12      12        1     |_ _ _ _|  _| |  _ _|_| | 7       8      4+4       2     |_ _ _ _| |_ _|_|    _ _| 8      15      15        1     |_ _ _ _ _|  _|     | 9      13      5+3+5     3     |_ _ _ _ _| |      _| 10     18      9+9       2     |_ _ _ _ _ _|  _ _| 11     12      6+6       2     |_ _ _ _ _ _| | 12     28      28        1     |_ _ _ _ _ _ _| ... For n = 9 the sum of divisors of 9 is 1+3+9 = A000203(9) = 13. On the other hand the 9th set of symmetric regions of the diagram is formed by three regions (or parts) with 5, 3 and 5 cells, so the total number of cells is 5+3+5 = 13, equaling the sum of divisors of 9. There are three parts: [5, 3, 5], so a(9) = 3. From _Omar E. Pol, Dec 21 2016: (Start) Illustration of the diagram of subparts (n = 1..12): --------------------------------------------------------- n   A000203  A279391  A001227           Diagram --------------------------------------------------------- .                               _ _ _ _ _ _ _ _ _ _ _ _ 1       1      1         1     |_| | | | | | | | | | | | 2       3      3         1     |_ _|_| | | | | | | | | | 3       4      2+2       2     |_ _|  _|_| | | | | | | | 4       7      7         1     |_ _ _|  _ _|_| | | | | | 5       6      3+3       2     |_ _ _| |_|  _ _|_| | | | 6      12      11+1      2     |_ _ _ _|  _| |  _ _|_| | 7       8      4+4       2     |_ _ _ _| |_ _|_|  _ _ _| 8      15      15        1     |_ _ _ _ _|  _|  _| | 9      13      5+3+5     3     |_ _ _ _ _| |  _|  _| 10     18      9+9       2     |_ _ _ _ _ _| |_ _| 11     12      6+6       2     |_ _ _ _ _ _| | 12     28      23+5      2     |_ _ _ _ _ _ _| ... For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2. For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End) From Hartmut F. W. Hoft, Dec 26 2016: (Start) Two examples of the general argument in the Comments section: Rows 27 in A237048 and A249223 (4 parts) i:  1  2 3 4 5 6 7 8 9 . . 12 27: 1  1 1 0 0 1                           1's in A237048 for odd divisors     1 27 3     9                           odd divisors represented 27: 1  0 1 1 1 0 0 1 1 1 0 1               blocks forming parts in A249223 Rows 81 in A237048 and A249223 (5 parts) i:  1  2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24 81: 1  1 1 0 0 1 0 0 1 0 0 0                          1's in A237048 f.o.d     1 81 3    27     9                                odd div. represented 81: 1  0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1  blocks fp in A249223 (End) MATHEMATICA a237271[n_] := Length[a237270[n]] (* code defined in A237270 *) Map[a237271, Range] (* data *) (* Hartmut F. W. Hoft, Jun 23 2014 *) PROG (PARI) fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell = i - 1; cell = j; vcells[ic] = cell; ic ++; ); ); ); vcells; } findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli == 0) && (vcelli == 0), return (i)); ); return (0); } findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli==x) && (vcelli==y) && (vcelli == 0) && (vcelli == 0), return (i)); ); return (0); } findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli == iz) && (vcelli == 0), return (i)); ); return (0); } zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i]); ); nbz; } docell(vcells, ic, iz) = {x = vcells[ic]; y = vcells[ic]; if (icdo = findxy(vcells, x-1, y), vcells[icdo] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo] = iz); vcells[ic] = 1; vcells; } docells(vcells, ic, iz) = {vcells[ic] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz); ); vcells; } nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++; ); zcount(vcells); } lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0); ); } \\ with heights() also defined in A237593; \\ Michel Marcus, Mar 28 2014 CROSSREFS Row lengths of A237270. Column 1 of A279387. - Omar E. Pol, Dec 16 2016 Cf. A000203, A001227, A024916, A060831, A061345, A175254, A196020, A235791, A236104, A237048, A237590, A237591, A237593, A239657, A244050, A244971, A245092, A249223, A250068, A261699, A262612, A262626, A279387, A279693. Sequence in context: A023120 A167970 A126433 * A176725 A085029 A185318 Adjacent sequences:  A237268 A237269 A237270 * A237272 A237273 A237274 KEYWORD nonn AUTHOR Omar E. Pol, Feb 25 2014 STATUS approved

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Last modified October 15 19:25 EDT 2019. Contains 328037 sequences. (Running on oeis4.)