A035324 A convolution triangle of numbers, generalizing Pascal's triangle A007318.
1, 3, 1, 10, 6, 1, 35, 29, 9, 1, 126, 130, 57, 12, 1, 462, 562, 312, 94, 15, 1, 1716, 2380, 1578, 608, 140, 18, 1, 6435, 9949, 7599, 3525, 1045, 195, 21, 1, 24310, 41226, 35401, 19044, 6835, 1650, 259, 24, 1, 92378, 169766, 161052, 97954, 40963, 12021, 2450
Offset: 1
Examples
Triangle begins:
1;
3, 1;
10, 6, 1;
35, 29, 9, 1;
126, 130, 57, 12, 1;
462, 562, 312, 94, 15, 1;
Triangle (0, 3, 1/3, 5/3, 3/5, ...) DELTA (1,0,0,0,0,0, ...) has an additional first column (1,0,0,...).
Links
- Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
- Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
- Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
- Wolfdieter Lang, First 10 rows.
Crossrefs
Programs
-
Haskell
a035324 n k = a035324_tabl !! (n-1) !! (k-1) a035324_row n = a035324_tabl !! (n-1) a035324_tabl = map snd $ iterate f (1, [1]) where f (i, xs) = (i + 1, map (`div` (i + 1)) $ zipWith (+) ((map (* 2) $ zipWith (*) [2 * i + 1 ..] xs) ++ [0]) ([0] ++ zipWith (*) [2 ..] xs)) -- Reinhard Zumkeller, Jun 30 2013 -
Mathematica
a[n_, m_] /; n >= m >= 1 := a[n, m] = 2*(2*(n-1) + m)*(a[n-1, m]/n) + m*(a[n-1, m-1]/n); a[n_, m_] /; n < m = 0; a[n_, 0] = 0; a[1, 1] = 1; Flatten[ Table[ a[n, m], {n, 1, 10}, {m, 1, n}]] (* Jean-François Alcover, Feb 21 2012, from first formula *) -
Sage
@cached_function def T(n, k): if n == 0: return n^k return sum(binomial(2*i-1, i)*T(n-1, k-i) for i in (1..k-n+1)) A035324 = lambda n,k: T(k, n) for n in (1..8): print([A035324(n, k) for k in (1..n)]) # Peter Luschny, Aug 16 2016
Formula
a(n+1, m) = 2*(2*n+m)*a(n, m)/(n+1) + m*a(n, m-1)/(n+1), n >= m >= 1; a(n, m) := 0, n
G.f. for column m: ((x*c(x)/sqrt(1-4*x))^m)/x, where c(x) = g.f. for Catalan numbers A000108.
a(n, m) =: s2(3; n, m).
With offset 0 (0 <= k <= n), T(n,k) = Sum_{j>=0} A039598(n,j)*binomial(j,k). - Philippe Deléham, Mar 30 2007
T(n+1,n) = 3*n = A008585(n).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + Sum_{i>=0} T(n-1,k+1+i)*(-1)^i. - Philippe Deléham, Feb 23 2012
T(n,m) = Sum_{k=m..n} k*binomial(k-1,k-m)*2^(k-m)*binomial(2*n-k-1,n-k)/n. - Vladimir Kruchinin, Aug 07 2013
A274230 Number of holes in a sheet of paper when you fold it n times and cut off the four corners.
0, 0, 1, 3, 9, 21, 49, 105, 225, 465, 961, 1953, 3969, 8001, 16129, 32385, 65025, 130305, 261121, 522753, 1046529, 2094081, 4190209, 8382465, 16769025, 33542145, 67092481, 134193153, 268402689, 536821761, 1073676289, 2147385345
Offset: 0
Comments
The folds are always made so the longer side becomes the shorter side.
We could have counted not only the holes but also all the notches: 4, 6, 9, 15, 25, 45, 81, 153, 289, ... which has the formula a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) and appears to match the sequence A183978. - Philippe Gibone, Jul 06 2016
The same sequence (0,0,1,3,9,21,49,...) turns up when you start with an isosceles right triangular piece of paper and repeatedly fold it in half, snipping corners as you go. Is there an easy way to see why the two questions have the same answer? - James Propp, Jul 05 2016
Reply from Tom Karzes, Jul 05 2016: (Start)
This case seems a little more complicated than the rectangular case, since with the triangle you alternate between horizontal/vertical folds vs. diagonal folds, and the resulting fold pattern is more complex, but I think the basic argument is essentially the same.
Note that with the triangle, the first hole doesn't appear until after you've made 3 folds, so if you start counting at zero folds, you have three leading zeros in the sequence: 0,0,0,1,3,9,21,... (End)
Also the number of subsets of {1,2,...,n} that contain both even and odd numbers. For example, a(3)=3 and the 3 subsets are {1,2}, {2,3}, {1,2,3}; a(4)=9 and the 9 subsets are {1,2}, {1,4}, {2,3}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}. (See comments in A052551 for the number of subsets of {1,2,...,n} that contain only odd and even numbers.) - Enrique Navarrete, Mar 26 2018
Also the number of integer compositions of n + 1 with an odd part other than the first or last. The complementary compositions are counted by A052955(n>0) = A027383(n) + 1. - Gus Wiseman, Feb 05 2022
Also the number of unit squares in the (n+1)-st iteration in the version of the dragon curve where the rotation directions alternate, so that any clockwise rotation is followed by a counterclockwise rotation, and vice versa (see image link below). - Talmon Silver, May 09 2023
Links
- Paolo P. Lava, Table of n, a(n) for n = 0..1000
- Philippe Gibone, Illustration of a(0)-a(4) (idealized).
- Talmon Silver, Illustration of alternating dragon.
- N. J. A. Sloane, Illustration for a(4) = 9 (scan of an actual cut-up piece of paper)
- N. J. A. Sloane, Illustration for a(5) = 21 (scan of an actual cut-up piece of paper)
- Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).
Programs
-
Magma
[(2^Ceiling(n/2)-1)*(2^Floor(n/2)-1 ): n in [0..35]]; // Vincenzo Librandi, Jul 02 2016
-
Maple
A274230:=n->(1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))): seq(A274230(n), n=0..50); # Wesley Ivan Hurt, Jul 07 2016
-
Mathematica
Table[(2^Ceiling@ # - 1) (2^Floor@ # - 1) &[n/2], {n, 0, 31}] (* Michael De Vlieger, Jun 30 2016 *) -
PARI
a(n)=2^n+1-(n%2+2)<<(n\2) \\ Charles R Greathouse IV, Jul 05 2016
Formula
u(0) = 0; v(0) = 0; u(n+1) = v(n); v(n+1) = 2u(n) + 1; a(n) = u(n)*v(n).
a(n) = (2^ceiling(n/2) - 1)*(2^floor(n/2) - 1).
Proof from Tom Karzes, Jul 05 2016: (Start)
Let r be the number of times you fold along one axis and s be the number of times you fold along the other axis. So r is ceiling(n/2) and s is floor(n/2), where n is the total number of folds.
When unfolded, the resulting paper has been divided into a grid of (2^r) by (2^s) rectangles. The interior grid lines will have diamond-shaped holes where they intersect (assuming diagonal cuts).
There are (2^r-1) internal grid lines along one axis and (2^s-1) along the other. The total number of internal grid line intersections is therefore (2^r-1)*(2^s-1), or (2^ceiling(n/2)-1)*(2^floor(n/2)-1) as claimed. (End)
From Colin Barker, Jun 22 2016, revised by N. J. A. Sloane, Jul 05 2016: (Start)
It follows that:
a(n) = (2^(n/2)-1)^2 for n even, a(n) = 2^n+1-3*2^((n-1)/2) for n odd.
a(n) = 3*a(n-1)-6*a(n-3)+4*a(n-4) for n>3.
G.f.: x^2 / ((1-x)*(1-2*x)*(1-2*x^2)).
a(n) = (1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))). (End)
a(n) = A079667(2^(n-1)) for n >= 1. - J. M. Bergot, Jan 18 2021
E.g.f.: cosh(x) + cosh(2*x) - 2*cosh(sqrt(2)*x) + sinh(x) + sinh(2*x) - 3*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, Apr 06 2022
A346633 Sum of even-indexed parts (even bisection) of the n-th composition in standard order.
0, 0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 3, 2, 1, 2, 0, 1, 2, 1, 3, 2, 1, 2, 4, 3, 2, 3, 1, 2, 3, 2, 0, 1, 2, 1, 3, 2, 1, 2, 4, 3, 2, 3, 1, 2, 3, 2, 5, 4, 3, 4, 2, 3, 4, 3, 1, 2, 3, 2, 4, 3, 2, 3, 0, 1, 2, 1, 3, 2, 1, 2, 4, 3, 2, 3, 1, 2, 3, 2, 5, 4, 3, 4, 2, 3, 4
Offset: 0
Keywords
Comments
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
Examples
Composition number 741 in standard order is (2,1,1,3,2,1), so a(741) = 1 + 3 + 1 = 5.
Crossrefs
Including odd-indexed parts gives A029837.
Subtracting from the odd version gives A124754.
Positions of zeros are A131577.
The odd-indexed version is A209281(n+1).
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A345197 counts compositions by sum, length, and alternating sum.
Programs
-
Mathematica
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse; Table[Total[Last/@Partition[Append[stc[n],0],2]],{n,0,100}]
A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.
1, 1, 0, 1, 3, 3, 6, 15, 24, 43, 92, 171, 315, 629, 1218, 2313, 4523, 8835, 17076, 33299, 65169
Offset: 0
Comments
The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. So a(n) is the number of compositions of n of length (n + s)/2, where s is the alternating sum of the composition.
Examples
The a(0) = 1 through a(7) = 15 compositions of n = 0..7 of length (n + s)/2 where s = alternating sum (empty column indicated by dot):
() (1) . (2,1) (2,2) (2,3) (2,4) (2,5)
(1,1,2) (1,2,2) (1,3,2) (1,4,2)
(2,1,1) (2,2,1) (2,3,1) (2,4,1)
(1,1,3,1) (1,1,3,2)
(2,1,2,1) (1,2,3,1)
(3,1,1,1) (2,1,2,2)
(2,2,2,1)
(3,1,1,2)
(3,2,1,1)
(1,1,1,1,3)
(1,1,2,1,2)
(1,1,3,1,1)
(2,1,1,1,2)
(2,1,2,1,1)
(3,1,1,1,1)
Crossrefs
Programs
-
Mathematica
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}]; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==(n+ats[#])/2&]],{n,0,15}]
A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.
0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131
Offset: 1
Keywords
Comments
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.
Examples
The terms together with their binary indices begin:
0: ()
9: (3,1)
11: (2,1,1)
14: (1,1,2)
130: (6,2)
133: (5,2,1)
135: (5,1,1,1)
138: (4,2,2)
141: (4,1,2,1)
143: (4,1,1,1,1)
148: (3,2,3)
153: (3,1,3,1)
155: (3,1,2,1,1)
158: (3,1,1,1,2)
Links
Crossrefs
Programs
-
Mathematica
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}]; stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse; Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]
A344743 Number of integer partitions of 2n with reverse-alternating sum < 0.
0, 0, 1, 3, 7, 15, 29, 54, 96, 165, 275, 449, 716, 1123, 1732, 2635, 3955, 5871, 8620, 12536, 18065, 25821, 36617, 51560, 72105, 100204, 138417, 190134, 259772, 353134, 477734, 643354, 862604, 1151773, 1531738, 2029305, 2678650, 3523378, 4618835, 6035240, 7861292
Offset: 0
Keywords
Comments
Conjecture: a(n) >= A236914.
The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i. This is equal to (-1)^(m-1) times the number of odd parts in the conjugate partition, where m is the number of parts. So a(n) is the number of even-length partitions of 2n with at least one odd conjugate part. By conjugation, this is also the number of partitions of 2n with greatest part even and at least one odd part.
The alternating sum of a partition is never < 0, so the non-reverse version is A000004.
Examples
The a(2) = 1 through a(5) = 15 partitions:
(31) (42) (53) (64)
(51) (62) (73)
(3111) (71) (82)
(3221) (91)
(4211) (3331)
(5111) (4222)
(311111) (4321)
(5221)
(5311)
(6211)
(7111)
(322111)
(421111)
(511111)
(31111111)
Crossrefs
Programs
-
Mathematica
sats[y_] := Sum[(-1)^(i - Length[y])*y[[i]], {i, Length[y]}]; Table[Length[Select[IntegerPartitions[n],sats[#]<0&]],{n,0,30,2}]
Extensions
More terms from Bert Dobbelaere, Jun 12 2021
A347449 Number of integer partitions of n with reverse-alternating product > 1.
0, 0, 1, 1, 2, 2, 5, 5, 10, 11, 20, 22, 37, 41, 66, 75, 113, 129, 190, 218, 310, 358, 497, 576, 782, 908, 1212, 1411, 1851, 2156, 2793, 3255, 4163, 4853, 6142, 7159, 8972, 10451, 12989, 15123, 18646, 21689, 26561, 30867, 37556, 43599, 52743, 61161, 73593
Offset: 0
Keywords
Comments
All such partitions have odd length.
We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)). The reverse-alternating product is the alternating product of the reversed sequence.
Examples
The a(2) = 1 through a(9) = 11 partitions:
(2) (3) (4) (5) (6) (7) (8) (9)
(211) (311) (222) (322) (332) (333)
(321) (421) (422) (432)
(411) (511) (431) (522)
(21111) (31111) (521) (531)
(611) (621)
(22211) (711)
(32111) (32211)
(41111) (42111)
(2111111) (51111)
(3111111)
Crossrefs
The strict case is A067659, except that a(0) = a(1) = 0.
The even bisection is A236559.
The case of >= 1 instead of > 1 is A344607.
The opposite version is A344608, also the non-reverse even-length case.
Allowing any integer reverse-alternating product gives A347445.
Allowing any integer alternating product gives A347446.
Reverse version of A347448; also the odd-length case.
The Heinz numbers of these partitions are the complement of A347450.
The multiplicative version (factorizations) is A347705.
A000041 counts partitions.
A027187 counts partitions of even length.
A027193 counts partitions of odd length.
A100824 counts partitions of n with alternating sum <= 1.
A347462 counts possible reverse-alternating products of partitions.
Programs
-
Mathematica
altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}]; Table[Length[Select[IntegerPartitions[n],altprod[Reverse[#]]>1&]],{n,0,30}]
A029760 A sum with next-to-central binomial coefficients of even order, Catalan related.
1, 8, 47, 244, 1186, 5536, 25147, 112028, 491870, 2135440, 9188406, 39249768, 166656772, 704069248, 2961699667, 12412521388, 51854046982, 216013684528, 897632738722, 3721813363288, 15401045060572, 63616796642368, 262357557683422, 1080387930269464
Offset: 0
Keywords
Comments
Proof by induction.
a(n) = total area below paths consisting of steps east (1,0) and north (0,1) from (0,0) to (n+2,n+2) that stay weakly below y=x. For example, the two paths with n=0 are
. _|.....|
The first has area 1 below it, the second area 0 and so a(0)=1. - David Callan, Dec 09 2004
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..1657
- Ran Pan, Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
- Sittipong Thamrongpairoj, Dowling Set Partitions, and Positional Marked Patterns, Ph. D. Dissertation, University of California-San Diego (2019).
Programs
-
Mathematica
a[n_] := (n+3)^2 CatalanNumber[n+2]/2 - 2^(2n+3); Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Sep 25 2018 *)
Formula
a(n) = 4^(n+1)*Sum_{k=1..n+1} binomial(2k, k-1)/4^k = ((n+3)^2)*C(n+2)/2-2^(2*n+3), C = Catalan. Also a(n+1)=4*a(n)+binomial(2(n+2), n+1).
G.f.: (d/dx)c(x)/(1-4*x), where c(x) = g.f. for Catalan numbers; convolution of A001791 and powers of 4. G.f. also c(x)^2/(1-4*x)^(3/2); convolution of Catalan numbers A000108 C(n), n >= 1, with A002457; convolution of A008549(n), n >= 1, with A000984 (central binomial coefficients).
a(n) = Sum_{k=0..n+1} A039598(n+1,k)*k^2. - Philippe Deléham, Dec 16 2007
A100824 Number of partitions of n with at most one odd part.
1, 1, 1, 2, 2, 4, 3, 7, 5, 12, 7, 19, 11, 30, 15, 45, 22, 67, 30, 97, 42, 139, 56, 195, 77, 272, 101, 373, 135, 508, 176, 684, 231, 915, 297, 1212, 385, 1597, 490, 2087, 627, 2714, 792, 3506, 1002, 4508, 1255, 5763, 1575, 7338, 1958, 9296, 2436, 11732, 3010, 14742
Offset: 0
Comments
From Gus Wiseman, Jan 21 2022: (Start)
Also the number of integer partitions of n with alternating sum <= 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These are the conjugates of partitions with at most one odd part. For example, the a(1) = 1 through a(9) = 12 partitions with alternating sum <= 1 are:
1 11 21 22 32 33 43 44 54
111 1111 221 2211 331 2222 441
2111 111111 2221 3311 3222
11111 3211 221111 3321
22111 11111111 4311
211111 22221
1111111 33111
222111
321111
2211111
21111111
111111111
(End)
Examples
From _Gus Wiseman_, Jan 21 2022: (Start)
The a(1) = 1 through a(9) = 12 partitions with at most one odd part:
(1) (2) (3) (4) (5) (6) (7) (8) (9)
(21) (22) (32) (42) (43) (44) (54)
(41) (222) (52) (62) (63)
(221) (61) (422) (72)
(322) (2222) (81)
(421) (432)
(2221) (441)
(522)
(621)
(3222)
(4221)
(22221)
(End)
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
-
Maple
seq(coeff(convert(series((1+x/(1-x^2))/mul(1-x^(2*i),i=1..100),x,100),polynom),x,n),n=0..60); (C. Ronaldo)
-
Mathematica
nmax = 50; CoefficientList[Series[(1+x/(1-x^2)) * Product[1/(1-x^(2*k)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *) Table[Length[Select[IntegerPartitions[n],Count[#,?OddQ]<=1&]],{n,0,30}] (* _Gus Wiseman, Jan 21 2022 *) -
PARI
a(n) = if(n%2==0, numbpart(n/2), sum(i=1, (n+1)\2, numbpart((n-2*i+1)\2))) \\ David A. Corneth, Jan 23 2022
Formula
G.f.: (1+x/(1-x^2))/Product(1-x^(2*i), i=1..infinity). More generally, g.f. for number of partitions of n with at most k odd parts is (1+Sum(x^i/Product(1-x^(2*j), j=1..i), i=1..k))/Product(1-x^(2*i), i=1..infinity).
a(n) ~ exp(sqrt(n/3)*Pi) / (2*sqrt(3)*n) if n is even and a(n) ~ exp(sqrt(n/3)*Pi) / (2*Pi*sqrt(n)) if n is odd. - Vaclav Kotesovec, Mar 07 2016
Extensions
More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005
A347465 Numbers whose multiset of prime indices has alternating product > 1.
3, 5, 7, 11, 12, 13, 17, 19, 20, 23, 27, 28, 29, 30, 31, 37, 41, 42, 43, 44, 45, 47, 48, 52, 53, 59, 61, 63, 66, 67, 68, 70, 71, 73, 75, 76, 78, 79, 80, 83, 89, 92, 97, 99, 101, 102, 103, 105, 107, 108, 109, 110, 112, 113, 114, 116, 117, 120, 124, 125, 127
Offset: 1
Keywords
Comments
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)).
All terms have odd bigomega (A001222).
Also Heinz numbers integer partitions with reverse-alternating product > 1.
Examples
The terms and their prime indices begin:
3: {2} 37: {12} 68: {1,1,7}
5: {3} 41: {13} 70: {1,3,4}
7: {4} 42: {1,2,4} 71: {20}
11: {5} 43: {14} 73: {21}
12: {1,1,2} 44: {1,1,5} 75: {2,3,3}
13: {6} 45: {2,2,3} 76: {1,1,8}
17: {7} 47: {15} 78: {1,2,6}
19: {8} 48: {1,1,1,1,2} 79: {22}
20: {1,1,3} 52: {1,1,6} 80: {1,1,1,1,3}
23: {9} 53: {16} 83: {23}
27: {2,2,2} 59: {17} 89: {24}
28: {1,1,4} 61: {18} 92: {1,1,9}
29: {10} 63: {2,2,4} 97: {25}
30: {1,2,3} 66: {1,2,5} 99: {2,2,5}
31: {11} 67: {19} 101: {26}
Crossrefs
Programs
-
Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}]; Select[Range[100],altprod[primeMS[#]]>1&]
Comments