A132273
a(n) = Sum{k=1..n} (k-th integer from among those positive integers that are coprime to (n+1-k)).
Original entry on oeis.org
1, 3, 7, 12, 20, 28, 41, 52, 69, 83, 103, 122, 149, 169, 197, 222, 257, 285, 322, 355, 397, 431, 477, 514, 567, 610, 662, 708, 769, 815, 882, 935, 1000, 1056, 1123, 1182, 1267, 1326, 1404, 1471, 1554, 1628, 1712, 1790, 1882, 1958, 2057, 2137, 2240
Offset: 1
The integers coprime to 1 are 1,2,3,4,5,6,... The 5th of these is 5. The integers coprime to 2 are 1,3,5,7,9,... The 4th of these is 7. The integers coprime to 3 are 1,2,4,5,7,... The 3rd of these is 4. The integers coprime to 4 are 1,3,5,... The 2nd of these is 3. And the integers coprime to 5 are 1,2,3,4,6,... The first of these is 1. So a(5) = 5 + 7 + 4 + 3 + 1 = 20.
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a132273 n = sum $ zipWith (!!) coprimess (reverse [0..n-1]) where
coprimess = map (\x -> filter ((== 1) . (gcd x)) [1..]) [1..]
-- Reinhard Zumkeller, Jul 08 2012
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a = {}; For[n = 1, n < 50, n++, s = 0; For[k = 1, k < n + 1, k++, c = 0; i = 1; While[c < k, If[GCD[i, n + 1 - k] == 1, c++ ]; i++ ]; s = s + i - 1]; AppendTo[a, s]]; a (* Stefan Steinerberger, Nov 01 2007 *)
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cop(k, j) = {my(nbc = 0, i = 0); while (nbc != j, i++; if (gcd(i,k)==1, nbc++)); i;}
a(n) = vecsum(vector(n, k, cop(k, n-k+1))); \\ Michel Marcus, Mar 14 2018
A132274
a(1)=1; a(n+1) = Sum_{k=1..n} (k-th integer from among those positive integers which are coprime to a(n+1-k)).
Original entry on oeis.org
1, 1, 3, 6, 10, 19, 27, 41, 51, 66, 78, 101, 119, 145, 167, 197, 219, 247, 272, 306, 335, 371, 403, 443, 477, 521, 559, 609, 647, 693, 737, 789, 834, 886, 940, 996, 1055, 1118, 1176, 1243, 1306, 1385, 1450, 1523, 1596, 1676, 1749, 1844, 1914, 2010, 2092, 2188
Offset: 1
The integers coprime to a(1)=1 are 1,2,3,4,5,6,... The 5th of these is 5. The integers coprime to a(2)=1 are 1,2,3,4,5... The 4th of these is 4. The integers coprime to a(3)=3 are 1,2,4,5,7,... The 3rd of these is 4. The integers coprime to a(4)=6 are 1,5,7,11,... The 2nd of these is 5. And the integers coprime to a(5)=10 are 1,3,7,9,11,... The first of these is 1. So a(6) = 5 + 4 + 4 + 5 + 1 = 19.
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A132274 := proc(n) option remember; local a,k,an1k,kcoud,c ; if n = 1 then 1; else a :=0 ; for k from 1 to n-1 do an1k := procname(n-k) ; kcoud := 0 ; for c from 1 do if gcd(c,an1k) = 1 then kcoud := kcoud+1 ; fi; if kcoud = k then a := a+c ; break; fi; od: od: a; fi; end: seq(A132274(n),n=1..60) ; # R. J. Mathar, Jul 20 2009
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A132274[n_] := A132274[n] = Module[{a, k, an1k, kcoud, c}, If[n == 1, 1, a = 0; For[k = 1, k <= n-1, k++, an1k = A132274[n-k]; kcoud = 0; For[c = 1, True, c++, If[GCD[c, an1k] == 1, kcoud++]; If[kcoud == k, a = a+c; Break[]]]]; a]];
Table[A132274[n], {n, 1, 60}] (* Jean-François Alcover, Jan 28 2024, after R. J. Mathar *)
A130802
a(1) = 1; a(n+1) = Sum_{k=1..n} (a(k)-th integer from among those positive integers which are coprime to (n+1-k)).
Original entry on oeis.org
1, 1, 2, 4, 9, 20, 47, 110, 260, 614, 1448, 3421, 8081, 19092, 45107, 106567, 251768, 594816, 1405285, 3320066, 7843851, 18531547, 43781846, 103437135, 244376187, 577352823, 1364029309, 3222597827, 7613573030, 17987504932, 42496516727, 100400469160
Offset: 1
The integers coprime to 5 are 1, 2, 3, 4, 6, ... The a(1)-th=1st of these is 1. The integers coprime to 4 are 1, 3, 5, ... The a(2)-th=1st of these is 1. The integers coprime to 3 are 1, 2, 4, 5, ... The a(3)-th=2nd of these is 2. The integers coprime to 2 are 1, 3, 5, 7, 9, ... The a(4)-th=4th of these is 7. And the integers coprime to 1 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... The a(5)-th=9th of these is 9. So a(6) = 1 + 1 + 2 + 7 + 9 = 20.
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with(numtheory): fc:= proc(t,p) option remember; local m, j, h, pp; if p=1 then t else pp:= phi(p); m:= iquo(t,pp); j:= m*pp; h:= m*p-1; while jAlois P. Heinz, Aug 05 2009
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fc[t_, p_] := fc[t, p] = Module[{m, j, h, pp}, If[p == 1, t, pp = EulerPhi[p]; m = Quotient[t, pp]; j = m pp; h = m p - 1; While[j < t, h++; If[GCD[p, h] == 1, j++]]; h]];
a[n_] := a[n] = If [n == 1, 1, Sum[fc[a[k], (n - k)], {k, 1, n - 1}]];
Array[a, 35] (* Jean-François Alcover, Nov 19 2020, after Alois P. Heinz *)
Showing 1-3 of 3 results.
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