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%I #23 Oct 15 2020 16:36:37
%S 38,26,95,71,59,103,67,224,176,175,151,115,232,184,303,219,254,170,
%T 146,264,204,180,144,405,309,321,261,428,368,284,296,571,511,475,379,
%U 600,612,444,538,466,406,1254,1050,763,727,732,516,996,1080,840,952,772,688,724,844,712,556,1488,1392,1336,1144
%N Irregular triangle read by rows: T(n, k) gives the sums of the members of the primitive period of the unsigned Schick sequences for the odd numbers from A333855.
%C For Schick's sequences see comments in A332439. In A333848 the sum for members of the primitive periods of the unsigned Schick sequences SBB(N, q0 = 1) (BB for Brändli and Beyne) for the odd numbers N from A333854 are given. (In Schick's book p is used instead of odd N >= 3, and in A333848 his B(p) = 1).
%C The length of row n is A135303(A333855(n)) (the B numbers for A333855(n)).
%C The corresponding gcd(T(n,k), 2*A333855(n)) values are given in A333851. They are used for the formula of the length of the Euler tours ET(A333855(n), q0_k), for k = 1, 2, ..., B(A333855(n)) based on the unsigned Schick sequences.
%D Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Bokos Druck, Zürich, 2003 (ISBN 3-9522917-0-6). Tables 3.1 to 3.10, for odd p = 3..113 (with gaps), pp. 158-166.
%H Gerold Brändli and Tim Beyne, <a href="https://arxiv.org/abs/1504.02757">Modified Congruence Modulo n with Half the Amount of Residues</a>, arXiv:1504.02757 [math.NT], 2016.
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/2008.04300">On the Equivalence of Three Complete Cyclic Systems of Integers</a>, arXiv:2008.04300 [math.NT], 2020.
%F T(n, k) = Sum_{j=1..A003558(A333855(n))} SBB(A333855(n), q0_k)_j, with the unsigned Schick sequence SBB(N, q0) for all used initial values q0 = q0_k for k = 1, 2, ..., A135303(A333855(n)) (B numbers >= 2).
%e The irregular triangle T(n, k) begins (here A(n) = A333855(n)):
%e n, A(n) \ k 1 2 3 4 5 6 7 8 9 ...
%e -------------------------------------------------------------
%e 1, 17: 38 26
%e 2, 31: 95 71 59
%e 3, 33: 103 67
%e 4, 41: 224 176
%e 5, 43: 175 151 115
%e 6, 51: 232 184
%e 7, 57: 303 219
%e 8, 63: 254 170 146
%e 9, 65: 264 204 180 144
%e 10, 73: 405 309 321 261
%e 11, 85: 428 368 284 296
%e 12, 89: 571 511 475 379
%e 13, 91: 600 612 444
%e 14, 93: 538 466 406
%e 15, 97: 1254 1050
%e 16, 99: 763 727
%e 17, 105: 732 516
%e 18, 109: 996 1080 840
%e 19, 113: 952 772 688 724
%e 20, 117: 844 712 556
%e 21, 119: 1488 1392
%e 22, 123: 1336 1144
%e 23, 127: 637 517 457 469 433 385 385 361 325
%e 24, 129: 649 469 469 385 397 361
%e 25, 133: 1374 1218 1026
%e 28, 137: 2456 2168
%e ...
%e --------------------------------------------------------------------------
%e n = 1, N = 17, B(17) = A135303((17-1)/2) = 2. In cycle notation:
%e SBB(17, q0_1) = (1, 15, 13, 9) and SBB(17, q0_2) = (3, 11, 5, 7), with sums
%e T(1, 1) = 1 + 15 + 13 + 9 = 38 and T(1, 2) = 26. (38 + 26 = 64 = A333848(8) .)
%o (PARI) RRS(n) = select(x->(((x%2)==1) && (gcd(n, x)==1)), [1..n]);
%o isok8(m, n) = my(md = Mod(2, 2*n+1)^m); (md==1) || (md==-1);
%o A003558(n) = my(m=1); while(!isok8(m, n) , m++); m;
%o B(n) = eulerphi(n)/(2*A003558((n-1)/2));
%o fmiss(rrs, qs) = {for (i=1, #rrs, if (! setsearch(qs, rrs[i]), return (rrs[i])););}
%o listb(nn) = {my(v=List()); forstep (n=3, nn, 2, my(bn = B(n)); if (bn >= 2, listput(v, n););); Vec(v);}
%o persum(n) = {my(bn = B(n)); if (bn >= 2, my(vn = vector(bn)); my(q=1, qt = List()); my(p = A003558((n-1)/2)); my(rrs = RRS(n)); for (k=1, bn, my(qp = List()); q = fmiss(rrs, Set(qt)); listput(qp, q); listput(qt, q); for (i=1, p-1, q = abs(n-2*q); listput(qp, q); listput(qt, q);); vn[k] = vecsum(Vec(qp));); return (vn););}
%o listas(nn) = {my(v = listb(nn)); vector(#v, k, persum(v[k]));} \\ _Michel Marcus_, Jun 13 2020
%Y Cf. A332439, A333848, A333854, A333851, A333855.
%K nonn,tabf
%O 1,1
%A _Wolfdieter Lang_, Jun 08 2020
%E Some terms were corrected by _Michel Marcus_, Jun 11 2010
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